##
## One Sample t-test
##
## data: Return
## t = 5.1, df = 14, p-value = 0.00008
## alternative hypothesis: true mean is greater than 6
## 95 percent confidence interval:
## 8.793 Inf
## sample estimates:
## mean of x
## 10.27
The Type I error for this problem would be suggesting the mutual fund performed better than the S&P 500 when it really doesn’t.
b.Specify the null and alternative hypothesis for this test.
\[ \begin{align*} H_{0}: \mu = 6\% \\ H_{a}: \mu > 6\% \end{align*} \]
\[\begin{align*} H_{0}: p = 25\% \\ H_{a}: p > 25\% \end{align*} \]
The normality assumption holds since \(n*p > 10\).
The results would be:
Test Stat = 0.49
p-value = 0.312
Assuming a 0.05 level of alpha, do not reject the null in favor of the alternative. There is not enough information to suggest that the proportion of customers who made a purchase is greater than 25%. It seems as if the promotion was not a success.
##
## Welch Two Sample t-test
##
## data: grocery_1 and grocery_2
## t = -1.9, df = 38, p-value = 0.07
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -11.2318 0.3652
## sample estimates:
## mean of x mean of y
## 91.53 96.97
a.Specify the null and alternative hypothesis for this test.
\[\begin{align*} H_{0}: \mu_{1} = \mu_{2} \\ H_{a}: \mu_{1} \neq \mu_{2} \end{align*} \]
The test-stat is -1.90 with a p-value of 0.065. Using an alpha of 0.05, do not reject the null in favor of the alternative.