Question

Problem 1 (6 pointe study of the amount 10 wee The manager of Fred s Grocery is making a money customers spend in the store. A sample of kda the tollowing amounta epens and 15 Saturday morning shoppers revealed Meekday $18.88 turday $11.54 24-76 Average of Weekdays: 50.347 Average of Saturday: 57.36 Standará Deviation of Reekday: 24.258 Standard Deviation of Saturday: 22.344 24.33 27.26 35.79 42.31 53.77 62.94 73.59 76.51 88.09 5.78 36.87 46.78 56.04 58.45 60.98 62.67 66.89 71-65 75.61 81.87 84-71 85.80 than weekday shoppera. Use the Student t Test and a 5 significance level. Owner Fred Snead contends that Saturday morning shoppers spend more a. State the null and alternate hypotheses HI : Ho: b. State the decision rule. c. Compute the value of the test statistic

Answer 1

Solution:

Let $\mu_{1}$ be the mean for Saturday

Let $\mu_{2}$ be the mean for Weekday

a. State the null and alternative hypotheses.

Answer:

$\boldsymbol{H_{0}:\mu_{1}=\mu_{2}}$

$\boldsymbol{H_{1}:\mu_{1}>\mu_{2}}$

b. State the decision rule.

Answer: To find the decision rule, we need to first find the right-tailed critical value at 0.05 significance level and for $\small df=n_{1}+n_{2}-2=10+15-2=23$ . Using the t distribution table, we have:

$\small t_{critical}=1.714$

Therefore, the decision rule is:

Reject the null hypothesis if $\small \boldsymbol{t>1.714}$

c. Compute the value of the test statistic.

Answer:

$\large t=\frac{\bar{x}_{1}-\bar{x}_{2}}{s_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}$

$\large s_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}}$

$\large =\sqrt{\frac{(15-1)22.344^{2}+(10-1)24.258^{2}}{15+10-2}}$

  $\large =23.1118$

$\large \therefore t=\frac{57.36-50.347}{23.1118\sqrt{\frac{1}{15}+\frac{1}{10}}}$

$\large =\frac{7.013}{9.4354}$

$\large =0.743$

Therefore, the test statistic is $\large \boldsymbol{t=0.743}$