# Function for t-test Critical Values, Confidence interval, t-statistic & p-value
# given a vector of descriptive variables, or data Inputs: cl = confidence level
# in decimal form data = either sample data in vector form, or descriptive
# variables in vector form. Descriptive variables are c(bar_{x},mu,s,n) Outputs:
# for data - a confidence interval for descriptive statistics: a tibble with
# Critical t values, a Confidence interval, the t-Statistic for bar_{x}, the
# pValue for bar_{x}
tCrit <- function(cl, data, tail, mu = 0) {
    f.pV.v <- function(x, data) {
        pv <- pt(x, df = data[4] - 1)
        return(pv)
    }
    f.pV.d <- function(x, data) {
        pv <- pt(x, df = length(data) - 1)
        return(pv)
    }
    
    f.tS.v <- function(data) {
        ts.v <- (data[1] - data[2])/(data[3]/sqrt(data[4]))
        return(ts.v)
    }
    f.TS.d <- function(data) {
        ts.d <- (mean(data) - mu)/(sd(data)/sqrt(length(data)))
        return(ts.d)
    }
    fcI.v <- function(cV, data) {
        ci.v <- c(data[2] - (qt(cV, data[4] - 1) * data[3]/sqrt(data[4])), data[2] + 
            (qt(cV, data[4] - 1) * data[3]/sqrt(data[4])))
        return(ci.v)
    }
    f.cI.d <- function(cV, data) {
        ci.d <- c(mean(data) - qt(cV, length(data) - 1) * sd(data)/sqrt(length(data)), 
            mean(data) + qt(cV, length(data) - 1) * sd(data)/sqrt(length(data)))
    }
    output <- function(tc, ci, ts, pv) {
        cValues <- tibble::tribble(~Variable, ~Values, "Low tCrit", tc[1], "High tCrit", 
            tc[2], "Conf_Int.Low", ci[1], "Conf_Int.Hi", ci[2], "T Statistic", ts, 
            "pValue", pv)
    }
    if (length(data) == 4) {
        if (tail == "2" | tail == 2) {
            cV <- (1 - cl)/2 + cl
            ci <- fcI.v(cV, data)
            ts <- f.tS.v(data)
            tc <- c(-1 * qt(cV, data[4] - 1), qt(cV, data[4] - 1))
            pv <- 2 * f.pV.v(-1 * abs(ts), data)
            cValues <- output(tc, ci, ts, pv)
        } else {
            if (tail == "L" | tail == "l") {
                cV <- (1 - cl)
                ci <- fcI.v(cV, data)
                ts <- f.tS.v(data)
                tc <- c(qt(cV, data[4] - 1), NA)
                pv <- f.pV.v(ts, data)
                cValues <- output(tc, ci, ts, pv)
            } else {
                if (tail == "R" | tail == "r") {
                  cV <- cl
                  ci <- fcI.v(cV, data)
                  ts <- f.tS.v(data)
                  tc <- c(NA, qt(cV, data[4] - 1))
                  pv <- 1 - f.pV.v(ts, data)
                  cValues <- output(tc, ci, ts, pv)
                }
            }
        }
    } else if (length(data) > 4 | length(data) < 4) {
        if (tail == "2" | tail == 2) {
            cV <- (1 - cl)/2 + cl
            ci <- f.cI.d(cV, data)
            ts <- f.TS.d(data)
            pv <- 2 * f.pV.d(-1 * abs(ts), data)
            tc <- c(-1 * qt(cV, length(data) - 1), qt(cV, length(data) - 1))
            cValues <- output(tc, ci, ts, pv)
        } else {
            if (tail == "L" | tail == "l") {
                cV <- (1 - cl)
                tc <- c(qt(cV, length(data) - 1), NA)
                ci <- f.cI.d(cV, data)
                ts <- f.TS.d(data)
                pv <- f.pV.d(ts, data)
                cValues <- output(tc, ci, ts, pv)
            } else {
                if (tail == "R" | tail == "r") {
                  cV <- cl
                  tc <- c(NA, qt(cV, length(data - 1)))
                  ci <- f.cI.d(cV, data)
                  ts <- f.TS.d(data)
                  pv <- 1 - f.pV.d(ts, data)
                  cValues <- output(tc, ci, ts, pv)
                  
                }
            }
        }
    } else {
        return(NA)
    }
    return(cValues)
}
# 2 Tailed t-test Inputs: cl = confidence level m1,m2 = means of sample 1 & 2
# s1,s2 = standard deviations of sample 1 & 2 n1,n2 = number of obs for sample 1
# & 2 m0 = Allows adjustment to the difference you want to test for, default 0
# var = Are the variances close enough to pool the samples?F=no, T=Yes, default F
# Output: tibble with statistic names in col1 '$variables' and values in col2
# '$values' See tibble labels for each output
t.2 <- function(cl, m1, m2, s1, s2, n1, n2, m0 = 0, var = F) {
    pV <- function(x, df) {
        pv <- 2 * pt(-abs(x), df)
    }
    cV <- (1 - cl)/2 + cl
    if (var == FALSE) {
        se <- sqrt((s1^2/n1) + (s2^2/n2))
        df <- ((s1^2/n1 + s2^2/n2)^2)/((s1^2/n1)^2/(n1 - 1) + (s2^2/n2)^2/(n2 - 1))
    } else {
        df <- n1 + n2 - 2
        sd <- sqrt(((n1 - 1) * s1^2 + (n2 - 1) * s2^2)/df)
        se <- sqrt((1/n1 + 1/n2)) * sd
    }
    t <- (m1 - m2 - m0)/se
    lc <- -1 * qt(cV, df)
    hc <- qt(cV, df)
    pv <- pV(t, df)
    t2 <- tibble::tribble(~Variables, ~Values, "meanDiff", m1 - m2, "SE", se, "lCrit", 
        lc, "hCrit", hc, "t", t, "pValue", pv)
    return(t2)
}

For the R code herein to work, it will require the following steps:
1. unzip the uploaded PPUA_5301__Homework_5.zip with R or unzipping program of your choice.
2. Run the ‘Call Data’ chunk and the GSS chunk below it.

# setwd('C:\\Users\\Stephen\\Documents\\Northeastern\\PPUA 5301 -
# Introduction to Computational Statistics\\Homework\\Homework
# 5\\PPUA_5301__Homework_5')
this.dir <- dirname("Holsenbeck_S_5.Rmd")
setwd(this.dir)
getwd()  #Check to see that it's the directory the file was opened from

## [1] "C:/Users/Stephen/Documents/Northeastern/Git/ppua5301/Homework 5"

knitr::read_chunk("GSS.r")

You hypothesize that the average person is smarter than Sarah Palin. You know her IQ is 100. You give an IQ test to 100 randomly selected people, and get a mean of 104 and standard deviation of 22. Please show your work for each question.

What is your null hypothesis?
What is your research hypothesis?
What is your test statistic?
Do you prefer a one-tailed or two-tailed test here, and why?
What is your alpha and threshold (t statistic) value or values for your rejection region? (Whatever alpha you prefer is fine, just be sure to state it and explain why you chose it.)
Can you reject the null under a one-tailed test?
Can you reject the null under a two-tailed test?
What is your 95% confidence interval?
What is the p-value for your test results?

You hypothesize that men and women have different skill levels in playing Tetris. To test this, you have 50 men and 50 women play the game in a controlled setting. The mean score of the men is 1124 with a standard deviation of 200 and the mean score for the women is 1245, also with a standard deviation of 200.

Are these scores statistically significantly different? Show your work.
Do you reject your hypothesis or the null? What do you conclude from this experiment?

You think drinking the night before an exam might help performance on the exam the next morning. To test this, you select 100 of your closest friends, and randomly get 50 of them drunk the night before the exam, which you denote the treatment group. The next day, the treatment group gets a mean of 78 with a standard deviation of 10 and the control group gets a 75 with a standard deviation of 5.

Does the evidence show that drinking helped exam performance?

1 Using data of your choosing (or using simulated data), use R to conduct the following tests, and explain the results you get:

A standard one-sample hypothesis test.
A difference-in-means test with independent samples.
A difference-in-means test with dependent samples (ie, a paired t-test).
Manually verify the results in (a) using the mean and sd as calculated by R (ie, you don’t have to manually calculate the mean or sd by hand!).

Homework 5

1

You hypothesize that the average person is smarter than Sarah Palin. You know her IQ is 100. You give an IQ test to 100 randomly selected people, and get a mean of 104 and standard deviation of 22. Please show your work for each question.

a.

What is your null hypothesis? \(H_0=\mu=100\)

b.

What is your research hypothesis? \(H_a=\mu>100\)

c.

What is your test statistic? Since there are 100 obs in the sample, the data is expected to follow a normal distribution, and the z-stat should be sufficient. \(z = \frac{x-\bar{x}}{s}\)
\(z = \frac{104-100}{22}\)
\(z = \frac{4}{22}\)
\(z = .1818\)
or if you’re looking for the t-statistic: \(t_{stat}=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\)
\(t_{stat}=\frac{104-100}{\frac{22}{\sqrt{100}}}\)
\(t_{stat}=1.818\)

zs <- function(x, mu, sd) {
    z <- (x - mu)/sd
    return(z)
}
(z <- zs(104, 100, 22))

## [1] 0.1818182

tCrit(0.95, c(104, 100, 22, 100), "r")

## # A tibble: 5 x 2
##       Variable        Value
##          <chr>        <dbl>
## 1   High tCrit   1.66039116
## 2 Conf_Int.Low  96.34713946
## 3  Conf_Int.Hi 103.65286054
## 4  T Statistic   1.81818182
## 5       pValue   0.03603016

d.

Do you prefer a one-tailed or two-tailed test here, and why?

A)A one-tailed test because of the question and the alternative hypothesis. In the question, the sample mean and st.dev are already given, and the mean is slightly higher than 100. A two-tailed test might have made more sense because the mean could have been < or > 100, but in this case we already know the sample mean. Additionally, the alternative hypothesis is that the average person is smarter, so we’re looking for the \(\bar{x}\) to be greater than \(\mu\) by some statistically significant measure.

e.

What is your alpha and threshold (t statistic) value or values for your rejection region? (Whatever alpha you prefer is fine, just be sure to state it and explain why you chose it.)

A)\(\alpha=.05\), I chose the .05 alpha level because it’s standard, and implies that the sample mean will have a \(\frac{1}{20}\) chance of being outside the expected range. Given the high standard deviation, it’s intuitively highly unlikely that the alternative hypothesis will be supported at the .01 level. The .1 alpha level allows for the conclusion to be inaccurate 1 in 10 times. The middle ground, or a .05 alpha level seems appropriate. The \(t_{crit}=1.72\). The wording in the question is confusing because t-statistic is usually defined by \(t_{stat}=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\) and it provides the value to compare to the \(t_{crit}\) to test the hypothesis. I am assuming based on the context of it’s appearance: “threshold (t statistic) value or values for your rejection region” that what’s being sought is the critical t value(s), which are provided below.

# For detail on tCrit or t.2 functons, see the Custom Fns chunk in the rmd
tCrit(0.95, c(104, 100, 22, 100), "r")

## # A tibble: 5 x 2
##       Variable        Value
##          <chr>        <dbl>
## 1   High tCrit   1.66039116
## 2 Conf_Int.Low  96.34713946
## 3  Conf_Int.Hi 103.65286054
## 4  T Statistic   1.81818182
## 5       pValue   0.03603016

f.

Can you reject the null under a one-tailed test?

A)If \(t_{stat}>t_{crit}\) then the null hypothesis can be rejected. The calculation below indicates the null can be rejected in a one tailed test at the 95% confidence level. \(t_{stat}=1.818>t_{crit}=1.66\)

tCrit(0.95, c(104, 100, 22, 100), "r")

## # A tibble: 5 x 2
##       Variable        Value
##          <chr>        <dbl>
## 1   High tCrit   1.66039116
## 2 Conf_Int.Low  96.34713946
## 3  Conf_Int.Hi 103.65286054
## 4  T Statistic   1.81818182
## 5       pValue   0.03603016

g.

Can you reject the null under a two-tailed test?

The upper \(t_{crit}\) is higher in a two tailed test, thus the test statistic no longer falls within the rejection region. With a two-tailed test we fail to reject the null hypothesis. \(t_{stat}=1.818<t_{crit}=1.984\)

tCrit(0.95, c(104, 100, 22, 100), "2")

## # A tibble: 6 x 2
##       Variable       Values
##          <chr>        <dbl>
## 1    Low tCrit  -1.98421695
## 2   High tCrit   1.98421695
## 3 Conf_Int.Low  95.63472271
## 4  Conf_Int.Hi 104.36527729
## 5  T Statistic   1.81818182
## 6       pValue   0.07206032

h.

What is your 95% confidence interval? \(\text{Confidence Interval}:CI=\bar{x}\pm t*\frac{s}{\sqrt{N}}\)
\(CI=100\pm 1.984*\frac{22}{\sqrt{100}}\)
\(CI=100\pm 4.3648\)

tCrit(0.95, c(104, 100, 22, 100), "2")

## # A tibble: 6 x 2
##       Variable       Values
##          <chr>        <dbl>
## 1    Low tCrit  -1.98421695
## 2   High tCrit   1.98421695
## 3 Conf_Int.Low  95.63472271
## 4  Conf_Int.Hi 104.36527729
## 5  T Statistic   1.81818182
## 6       pValue   0.07206032

i.

What is the p-value for your test results?

A)It seems like everyone is saying something different about how a \(p_{value}\) relates to one v two tailed tests. Based on the information in this tutorial, the \(p_{value}\) is different depending on the test. IE for a 2 tailed test it is: \(p_{value}=2*\int\limits^{x}_{{-}\infty}f(-|t_{stat}|)dt\)
for a Left tailed test it is: \(p_{value}=\int\limits^{x}_{{-}\infty}f(t_{stat})dt\)
and for a Right tailed test it is: \(p_{value}=1-\int\limits^{x}_{{-}\infty}f(t_{stat})dt\)
Based on these formulas, the \(p_{value}\) is .03 for the right tailed test, rejecting the null hypothesis, and is .07 for the 2 tailed test, failing to reject the null hypothesis.

tCrit(0.95, c(104, 100, 22, 100), "2")

## # A tibble: 6 x 2
##       Variable       Values
##          <chr>        <dbl>
## 1    Low tCrit  -1.98421695
## 2   High tCrit   1.98421695
## 3 Conf_Int.Low  95.63472271
## 4  Conf_Int.Hi 104.36527729
## 5  T Statistic   1.81818182
## 6       pValue   0.07206032

2.

You hypothesize that men and women have different skill levels in playing Tetris. To test this, you have 50 men and 50 women play the game in a controlled setting. The mean score of the men is 1124 with a standard deviation of 200 and the mean score for the women is 1245, also with a standard deviation of 200.

a.

Are these scores statistically significantly different? Show your work.

Since the standard deviations are identical, the samples can be pooled, thus we first solve for the pooled standard deviation: \(s_p=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}\)
\(s_p=\sqrt{\frac{(50-1)200^2+(50-1)200^2}{50+50-2}}\)
\(s_p=\sqrt{40000}\)
\(s_p=200\)
Probably should have guessed 200 would be the \(s_p\) with identical \(n\) and \(s\) in the two samples. The next step is to solve for the \(t_{stat}\)
\(t_{stat}=\frac{\bar{x}_1-\bar{x}_2}{s_{p}*\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\)
\(t_{stat}=\frac{1124-1245}{200*\sqrt{\frac{1}{50}+\frac{1}{50}}}\)
\(t_{stat}=\frac{-121}{200*.2}\)
\(t_{stat}=-3.025\)
Then find the degrees of freedom to establish the \(t_{crit}\) \(df_{s1=s2}=n_1+n_2-2\)
\(df_{s1=s2}=50+50-2\)

t.2(0.95, 1124, 1245, 200, 200, 50, 50, var = T)

## # A tibble: 6 x 2
##   Variables        Values
##       <chr>         <dbl>
## 1  meanDiff -1.210000e+02
## 2        SE  4.000000e+01
## 3     lCrit -1.984467e+00
## 4     hCrit  1.984467e+00
## 5         t -3.025000e+00
## 6    pValue  3.174979e-03

b.

Do you reject your hypothesis or the null? What do you conclude from this experiment?

A)At the 95% confidence level the null hypothesis can be rejected in favor of the research hypothesis. It can be concluded that at the 95% confidence level there is a statistically signicant difference between the Tetris skills of males v females.

3.

You think drinking the night before an exam might help performance on the exam the next morning. To test this, you select 100 of your closest friends, and randomly get 50 of them to drink the night before the exam, which you denote the treatment group. The next day, the treatment group gets a mean of 78 with a standard deviation of 10 and the control group gets a 75 with a standard deviation of 5.

a.

Does the evidence show that drinking helped exam performance?

A)In this example, the standard deviations are different enough where it would be misleading to pool them. We can first compute the \(t_{stat}\): \(t_{stat}=\frac{{\bar{x}}_1-{\bar{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\)
\(t_{stat}=\frac{78-75}{\sqrt{\frac{100}{50}+\frac{25}{50}}}\)
\(t_{stat}=\frac{3}{1.581139}\)
\(t_{stat}=1.897367\)
and then the degrees of freedom: \(df=\frac{(n_1-1)*(n_2-1)}{(n_2-1)C^2+(1-C)^2(n_1-1)}\text{ where } C=\frac{s_1^2/n_1}{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\)
\(C=\frac{100/50}{\frac{100}{50}+\frac{25}{50}}\)
\(C=\frac{2}{2.5}\)
\(C=0.8\) \(df=\frac{(49)*(49)}{(49).8^2+(1-.8)^2(49)}\)
\(df=\frac{2401}{33.32}\)
\(df=72.059\)
With this degree of freedom the critical value is 1.993436, the \(t_{stat}=1.897367\) and does not fall within the rejection region, so at the 95% confidence level we fail to reject the null hypothesis and conclude that drinking before an exam does not improve exam performance.

# tstat
(ts <- 3/sqrt(100/50 + 25/50))

## [1] 1.897367

# df
(c <- 2/2.5)

## [1] 0.8

(df <- 49 * 49/((49) * 0.8^2 + (1 - 0.8)^2 * (49)))

## [1] 72.05882

# Critical Value
qt(0.975, df)

## [1] 1.993436

t.2(0.95, 78, 75, 10, 5, 50, 50, var = F)

## # A tibble: 6 x 2
##   Variables      Values
##       <chr>       <dbl>
## 1  meanDiff  3.00000000
## 2        SE  1.58113883
## 3     lCrit -1.99343576
## 4     hCrit  1.99343576
## 5         t  1.89736660
## 6    pValue  0.06178651

4.

Using data of your choosing (or using simulated data), use R to conduct the following tests, and explain the results you get: ### a. A standard one-sample hypothesis test.

data <- rt(100, 99)
t.test(data)

## 
##  One Sample t-test
## 
## data:  data
## t = 0.018249, df = 99, p-value = 0.9855
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  -0.1703142  0.1734761
## sample estimates:
##  mean of x 
## 0.00158092

A)100 random values within the t-distribution are stored to the “data” variable. Data is passed to the t.test function. The function, defaults to assume that \(H_0\) is that \(\mu=0\) and the \(H_a\) is \(\mu \neq0\) and that the confidence level is 95%. It performs the necessary calculations and returns the \(t_{stat},df.p_{value},H_a,\text{ a confidence interval, and } \bar{x}\). It’s clear that with a \(p_{value}\) of .8683, that we fail to reject the null and that the mean is close to 0. We can see that \(\bar{x}=0.015\) which is close to 0.

b.

A difference-in-means test with independent samples.

View(GSS)
# Variable HEALTH Question: Would you say your own health, in general, is
# excellent, good, fair, or poor Answers in tibble below
tibble::tribble(~Value, ~Answer, 1L, "Excellent", 2L, "Good", 3L, "Fair", 4L, "Poor", 
    8L, "Don't know", 9L, "No answer", 0L, "Not applicable")

## # A tibble: 7 x 2
##   Value         Answer
##   <int>          <chr>
## 1     1      Excellent
## 2     2           Good
## 3     3           Fair
## 4     4           Poor
## 5     8     Don't know
## 6     9      No answer
## 7     0 Not applicable

(Total.Resp <- GSS %>% group_by(YEAR) %>% summarise(n = n()))

## # A tibble: 3 x 2
##    YEAR     n
##   <int> <int>
## 1  2012  1974
## 2  2014  2538
## 3  2016  2867

(health_by_yr <- GSS %>% group_by(YEAR) %>% filter(HEALTH > 0 & HEALTH < 5) %>% summarise(mHealth = mean(HEALTH), 
    sHealth = sd(HEALTH), nHealth = n(), p.Resp = ))

## # A tibble: 3 x 4
##    YEAR  mHealth   sHealth nHealth
##   <int>    <dbl>     <dbl>   <int>
## 1  2012 2.069678 0.8535072    1306
## 2  2014 2.089474 0.8638472    1710
## 3  2016 2.131565 0.8266308    1885

health_by_yr <- cbind(health_by_yr, total.n = Total.Resp$n)
health_by_yr %>% mutate(p.Resp = nHealth/total.n)

##   YEAR  mHealth   sHealth nHealth total.n    p.Resp
## 1 2012 2.069678 0.8535072    1306    1974 0.6616008
## 2 2014 2.089474 0.8638472    1710    2538 0.6737589
## 3 2016 2.131565 0.8266308    1885    2867 0.6574817

health_by_yr12 <- GSS %>% group_by(YEAR) %>% filter(HEALTH > 0 & HEALTH < 5, YEAR == 
    2012)
health_by_yr16 <- GSS %>% group_by(YEAR) %>% filter(HEALTH > 0 & HEALTH < 5, YEAR == 
    2016)
t.test(health_by_yr12$HEALTH, health_by_yr16$HEALTH)

## 
##  Welch Two Sample t-test
## 
## data:  health_by_yr12$HEALTH and health_by_yr16$HEALTH
## t = -2.04, df = 2748.3, p-value = 0.04144
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.121370917 -0.002402242
## sample estimates:
## mean of x mean of y 
##  2.069678  2.131565

A)\(H_0=\mu=0\) The average respondents self-reported health is not different between years 2012 and 2016}
\(H_a=\mu\neq{0}\) The average respondents self-reported health is different between years 2012 and 2016} The 2 sample t.test \(p_{value}\) indicates that we reject the null and that the average response of responding participants (2012~66% & 2016~66%) is different at the 95% confidence level. We conclude that at the 95% confidence level, the average respondent indicates that their health is different, and better in 2016 than in 2012.

c.

A difference-in-means test with dependent samples (ie, a paired t-test).

A)The group will be respondents to the following question whom answered as having a “General Sense”" or “Clear Understanding”:

SCISTUDY: Now, for a slightly different type of question. When you read news stories, you see certain sets of words and terms. We are interested in how many people recognize certain kinds of terms. First, some articles refer to the results of a scientific study. When you read or hear the term scientific study, do you have a clear understanding of what it means, a general sense of what it means, or little understanding of what it means?

And we will compare their answers to the following questions:

ODD1: Now, think about this situation. A doctor tells a couple that their genetic makeup means that they have got one in four chances of having a child with an inherited illness. A. Does this mean that if their first child has the illness, the next three will not have the illness?

EARTHSUN: Now, I would like to ask you a few short questions like those you might see on a television game show. for each statement that I read, please tell me if it is true or false. If you don’t know or aren’t sure, just tell me so, and we will skip to the next question. Remember true, false, or don’t know. J. Now, does the Earth go around the Sun, or does the Sun go around the Earth?

The t-test is framed as follows:
\(H_0=\bar{x}_1-\bar{x}_2=0\) People who rate themselves as having a “General Sense” or “Clear Understanding” of scientific conclusions, will answer the question about probability and about the orbit of the earth around the sun with the same frequency of wrong and right answers. \(H_a=\bar{x}_1-\bar{x}_2\neq0\) People who rate themselves as having a “General Sense” or “Clear Understanding” of scientific conclusions, will answer the question about probability and about the orbit of the earth around the sun with different frequencies of wrong and right answers.

# Variable: SCISTUDY
SCISTUDY <- tibble::tribble(~Code, ~Label, 1L, "Clear.understanding", 2L, "General sense", 
    3L, "Little understanding", 8L, "Dont know", 9L, "No answer", 0L, "Not applicable")
# Variable: ODDS1
ODDS1 <- tibble::tribble(~Code, ~Label, 1L, "Yes", 2L, "No", 8L, "Dont know", 9L, 
    "No answer", 0L, "Not applicable")
# Variable: EARTHSUN
Earthsun <- tibble::tribble(~Code, ~Label, 1L, "Earth around sun", 2L, "Sun around earth", 
    8L, "Dont know", 9L, "No answer", 0L, "Not applicable")
(sci.s <- GSS %>% filter(SCISTUDY > 0) %>% group_by(ID_) %>% select(ID_, SCISTUDY, 
    ODDS1, EARTHSUN))

## # A tibble: 3,631 x 4
## # Groups:   ID_ [2,289]
##      ID_ SCISTUDY ODDS1 EARTHSUN
##    <int>    <int> <int>    <int>
##  1     2        2     2        1
##  2     3        2     2        1
##  3     4        2     2        1
##  4     5        1     2        1
##  5     8        8     2        1
##  6    10        1     2        1
##  7    13        1     2        1
##  8    15        2     1        1
##  9    16        2     1        1
## 10    17        2     8        1
## # ... with 3,621 more rows

switch <- function(x) {
    if (x == 1) {
        x <- 2
    } else {
        x <- 1
    }
}  # For switching answer 1 and 2 on EARTHSUN such that 1 is the wrong answer, as in ODDS1, so that the data can be compared via the paired t-test
sci.hi <- sci.s %>% filter((SCISTUDY == 1 | SCISTUDY == 2) & (EARTHSUN == 1 | EARTHSUN == 
    2) & (ODDS1 == 1 | ODDS1 == 2)) %>% mutate(ES = switch(EARTHSUN))
t.test(sci.hi$ODDS1, sci.hi$ES, paired = T)

## 
##  Paired t-test
## 
## data:  sci.hi$ODDS1 and sci.hi$ES
## t = 10.307, df = 2581, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  0.07965883 0.11708788
## sample estimates:
## mean of the differences 
##              0.09837335

t.test(sci.hi$ODDS1, sci.hi$ES)  #See the actual means

## 
##  Welch Two Sample t-test
## 
## data:  sci.hi$ODDS1 and sci.hi$ES
## t = 10.129, df = 4769.9, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  0.07933338 0.11741332
## sample estimates:
## mean of x mean of y 
##  1.903950  1.805577

At the 95% Confidence level we can reject the null hypothesis in favor of the alternative, and show that the evidence supports the hypothesis that eople who rate themselves as having a “General Sense” or “Clear Understanding” of scientific conclusions, answer the question about probability and about the orbit of the earth around the sun with different frequencies of wrong and right answers. (The means in the unpaired t-test show that the mean for earthsun is closer to 1, indicating that people actually get the question about the orbit of the earth around the sun wrong with a higher frequency.)

d.

Manually verify the results in (a) using the mean and sd as calculated by R (ie, you don’t have to manually calculate the mean or sd by hand!).

First we can calculate the \(t_{stat}\)
\(t_{stat}=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\)
\(t_{stat}=\frac{0.0016-0}{\frac{0.8663}{\sqrt{100}}}\)
\(t_{stat}=\frac{0.0016}{.0866}\)
\(t_{stat}=0.01847\)
Given the \(t_{crit}\) is ~1.984, we can see that the \(t_{stat}<t_{crit}\) of the and thus the sample mean is nowhere near close to significantly different than the expected mean. Considering the sample data is a random distribution of 100 numbers on the t-distributon, we would hope this is the case.

mean(data)

## [1] 0.00158092

sd(data)

## [1] 0.8663122

0.0016/0.0866

## [1] 0.01847575

qt(0.975, 99)

## [1] 1.984217

tCrit(0.95, data, "2")  #Check the answers

## # A tibble: 6 x 2
##       Variable      Values
##          <chr>       <dbl>
## 1    Low tCrit -1.98421695
## 2   High tCrit  1.98421695
## 3 Conf_Int.Low -0.17031422
## 4  Conf_Int.Hi  0.17347606
## 5  T Statistic  0.01824885
## 6       pValue  0.98547706

Holsenbeck_S_5

Stephen Synchronicity

2017-10-12

Homework 5

1

a.

b.

c.

d.

e.

f.

g.

h.

i.

2.

a.

b.

3.

a.

4.

b.

c.

d.