Problem 1

Generate an object in R (call it whatever you want to name it) with 100 random draws from a standard normal distribution. Using this object, calculate and provide

set.seed(11)
random_numbers <- rnorm(100)
random_numbers
##   [1] -0.59103110  0.02659437 -1.51655310 -1.36265335  1.17848916 -0.93415132
##   [7]  1.32360565  0.62491779 -0.04572296 -1.00412058 -0.82843324 -0.34835173
##  [13] -1.53829340 -0.25556525 -1.14994503  0.01232697 -0.22296954  0.88777165
##  [19] -0.59215528 -0.65571812 -0.68251762 -0.01585819 -0.44260479  0.35255750
##  [25]  0.07317058  0.00715880 -0.18760011 -0.76570065 -0.22105682 -0.98358859
##  [31] -1.10428404 -0.93815021  0.67862424 -1.57749787 -0.86993846  0.48467705
##  [37] -0.18605270  1.54555470 -0.61138007 -0.34775649 -1.63651631  0.02038144
##  [43]  0.89174268 -0.87274968  0.89005083 -0.34387435 -2.18678137  0.88005818
##  [49]  0.72385656  0.21985268  0.78987057 -0.22999390 -0.81850248  0.49973416
##  [55]  0.15919235  0.54262643 -0.15664505  0.43879332  1.48787060  0.06016510
##  [61] -0.84901287  2.33969306 -0.12120295 -1.95020737  0.53871149  1.69351482
##  [67] -0.79096822 -1.07526060 -0.60787512  0.75440166  0.45347606 -0.12343368
##  [73] -0.76309682  0.22827009  1.11946192  0.15657318 -0.68877213  0.45294960
##  [79] -1.06754665  0.40156515 -0.06477369  0.31549629 -0.60568155 -0.90758469
##  [85]  2.26160898 -0.60322671 -1.29786210  0.50645120 -0.85333426 -1.50603179
##  [91]  1.20232890 -1.02786539  0.93826996 -0.54315466  0.51309513 -0.35259088
##  [97]  1.32653329 -1.14015185  1.41310994 -0.60217878

1a) A histogram of the object

hist(random_numbers,
     main = "Histogram of Random Normal Values",
     xlab = "Values",
     col = "purple")

1b) The mean

mean <- mean(random_numbers)
mean
## [1] -0.1235137

1c) The standard deviation

sd <- sd(random_numbers)
sd
## [1] 0.9144671

1d) The standard error of the mean

sd(random_numbers)/sqrt(length(random_numbers))
## [1] 0.09144671

1e) The 95% confidence interval

mean(random_numbers) + c(-1,1) * 1.96 * (sd(random_numbers)/sqrt(length(random_numbers)))
## [1] -0.30274928  0.05572183

Problem 2

Using the data (from the assignment sheet) to conduct a difference of means test:

Please make sure to show your work. Remember, you can use R as a calculator (as shown below) but you are also able to submit written work as well.

2a) Calculate the t-statistic

t_stat <- (2.0- 2.64) /
  sqrt((1.32^2 / 9) + (1.75^2 / 11))

t_stat
## [1] -0.931547

The t-statistic is…-0.931547

2b) Calculate the degrees of freedom

df <- (9)+(11)-(2)
df
## [1] 18

There are 18 degrees of freedom.

2c) Compare the t-statistic to the critical value for p = 0.05 for the corresponding degrees of freedom.

Written answer here…. 2.101>|-.92| so because of that we fail to reject the null hypothsis there is no difference statisticaly in groups with a lot of females vs a little females