First, we need to calculate the mean, \(\bar{x}\), by taking the average of the limits of the confidence interval:
\(\bar{x} = \frac{\text{Lower Limit }+\text{ Upper Limit}}{2} = \frac{18.985 + 21.015}{2} = \boxed{20}\)
We also know that the margin of error can be calculated with the following equation:
\(E = t^{*}\frac{s}{\sqrt{n}}\)
which can be rearranged to:
\(s = \frac{E\sqrt{n}}{t^{*}}\)
The margin of error, \(E\), can be calculated by the difference of the boundaries of the confidence interval divided by 2:
\(E = \frac{\text{Upper Limit }-\text{ Lower Limit}}{2} = \frac{21.015 - 18.985}{2} = 1.015\)
The sample size, \(n\), is 36, and the critical value, \(t^{*}\), can be found in the \(t\)-distribution chart with the degrees of freedom, \(df\), being equal to 35 (\(n-1\)), and the cutoff for the tail being 0.05 (\(1-\text{Confidence Level}\)). Or you can use the following R-code:
conf <- 0.95
samp_size <- 36
crit_t <- abs(qt((1-conf)/2,samp_size-1)) # Divide by 2 for two-tailed distribution
crit_t[1] 2.030108Now, we can pluh back in to the previous equation for standard deviation:
\(s = \frac{E\sqrt{n}}{t^{*}} = \frac{1.015\sqrt{36}}{2.030108} = 2.99984 \approx \boxed{3}\)
\(\boxed{\bar{x} = 20, s = 3}\)
We are given the following parameters:
\(s = \$100, E = \$10, \text{confidence} = 95\%\)
If the sample is greater than 30, we can assume normality and use the following equation for margin of error:
\(E = z^{*}\frac{s}{\sqrt{n}}\)
We can rearrange this equation to get:
\(n = \left(z^{*}\frac{s}{E}\right)^{2}\)
In order to get the critical z value, \(z^{*}\), we can look in the table or use R:
conf <- 0.95
tail <- (1-conf)/2 # Divide by 2 for two-tailed distribution
crit_z <- qnorm(tail,lower.tail=FALSE)
crit_z[1] 1.959964Now, we can plug this back into the equation and get the sample size:
\(n = \left(z^{*}\frac{s}{E}\right)^{2} = \left(1.959964\frac{100}{10}\right)^2 = 384.146\)
We need to round up to ensure enough people, and we get \(\boxed{n = 385}\).
We are given the following parameters:
\(n = 51, \bar{x}_{diff} = 1.1, s_{diff} = 4.9\)
Yes, there is a relationship. Both datasets contain the same 51 locations.
The null hypothesis for this would be that there is no difference between the temperatures in 1968 and 2008, and the alternative hypothesis would be that the temperature in 2008 is greater than 1968, which means the difference would be positive:
\(H_{0}: \mu_{diff} = 0\)
\(H_{a}: \mu_{diff} > 0\)
Yes. We can assume independence because the sample size is greater than 10. It’s random since all the locations were picked randomly. Finally, it’s normal because 50 degrees of freedom is shown to align almost exactly with a normal distribution.
The \(t\) value can be calculated by the following equation:
\(t = \frac{\bar{x}_{diff}}{s_{diff}\sqrt{n}} = \frac{1.1}{4.9/\sqrt{51}}\)
The p-value for the above t-statistic and a sample size of 51 can be calculated using the following R-code (assumed 95% confidence):
xd <- 1.1
sd <- 4.9
size <- 51
t <- xd/(sd/sqrt(size))
df <- size-1
pt(t,df,lower=FALSE)[1] 0.05759731Since the p-value is greater than 0.05, we failed to reject the null hypothesis, and there is not enough evidence to say that the temperature is warmer in 2008 than in 1968.
It is possible that we made a Type 2 Error, which is failing to reject the null hypothesis, when the null hypothesis is false. In our case, the temperature may actually be different, but we did not see that due to the error.
Since we failed to reject the null hypothesis, we should see 0 in the confidence interval because 0 may be the true average difference.
Even though the box plots contain different means, we can say that their distributions have approximately the same shape. Both sets have the means centered in the box, and the box centered in the whiskers. In addition, both sets have no outliers. Therefore, we can say that both follow a normal distribution.
Let’s denote the horsebean data with the subscript “1”, and denote the linseed data with the subscript “2”. The givens for both are:
\(\text{horsebean}: n_{1} = 10, \bar{x}_{1} = 160.20, s_{1} = 38.63\)
\(\text{linseed}: n_{2} = 12, \bar{x}_{2} = 218.75, s_{2} = 52.24\)
The problem also stated that they were looking for a difference between the two. Therefore, the hypotheses look like this:
\(H_{0}: \mu_{1} = \mu_{2}\)
\(H_{a}: \mu_{1} \ne \mu_{2}\)
The final given was that we are assuming a 5% significance level (\(\alpha=0.05\)). The equation to calculate the t-statistic is:
\(t = \frac{\bar{x}_{1}-\bar{x}_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}\)
The degrees of freedom, \(df\), is the sample size of the smaller sample minus 1. We can calulate the p-value with this information in R:
x1 <- 160.20
s1 <- 38.63
n1 <- 10
x2 <- 218.75
s2 <- 52.24
n2 = 12
vec1 = c(x1)
vec2 = c(x2)
t <- abs(x1-x2)/(sqrt(((s1^2)/n1)+((s2^2)/n2)))
df <- min(n1-1,n2-1)
2*pt(t,df,lower=FALSE)[1] 0.01455232Since the p-value is less than 0.05, we can reject the null hypothesis in favor of the alternative. Therefore, there is significant evidence that the means are different.
Since we rejected the null hypothesis, it is possible that we made a Type 1 Error. This means that we rejected the null hypothesis when it is actually true.
Yes, the conclusion would change since the p-value is greater than \(\alpha\). In this case, we would not reject the null hypothesis, and say that we do not have sufficient evidence to say the means are different.
\(H_{0}: \mu_{1} = \mu_{2} = \mu_{3} = \mu_{4} = \mu_{5}\)
\(H_{a}: \text{Not all means are equal}\)
The subscripts represent each of the groups.
This study is independent because the people in the different groups are independent. We can say that it is normally distributed because there are a lot of samples per group. We can also say that there are equal variances since the ratio of the largest standard deviation to the smallest is less than 2.
First, we need to calculate the degrees of freedom for the coffee groups, \(df_{G}\), residuals, \(df_{E}\), and the total, \(df_{T}\). The degrees of freedom are calculated below:
\(df_{G} = \text{Num. Groups} - 1 = 5 - 1 = \boxed{4}\)
\(df_{E} = n - \text{Num. Groups} = 50,739 - 5 = \boxed{50,734}\)
\(df_{T} = df_{C} + df_{R} = 4 + 50,734 = \boxed{50,738}\)
The Sum of the Squares for the coffee groups, \(SSG\), can be seen below:
\(SSG = SST - SSE = 25,575,327 - 25,564,819 = \boxed{10,508}\)
The Mean Square of the coffee groups, \(MSG\), and the Mean Square of the Residuals, \(MSE\), can be seen below:
\(MSG = \frac{SSG}{df_{G}} = \frac{10,508}{4} = \boxed{2,627}\)
\(MSE = \frac{SSE}{df_{E}} = \frac{25,564,819}{50734} = \boxed{503.899}\)
The \(F\)-value can be calculated by the following:
\(F = \frac{MSG}{MSE} = \frac{2,627}{503.899} = \boxed{5.21334}\)