Math  /  Data & Statistics

QuestionFree dessert: In an attempt to increase business on Monday nights, a restaurant offers a free dessert with every dinner order. Before the offer, the mean number of dinner customers on Monday was 150. Following are the numbers of diners on a random sample of 12 days while the offer was in effect. Can you conclude that the mean number of diners decreased while the free dessert offer was in effect? Use the α=0.01\alpha=0.01 level of significance and the PP-value method with the - Critical Values for the Student's t Distribution Table. \begin{tabular}{llllll} \hline 170 & 133 & 150 & 111 & 171 & 103 \\ 101 & 110 & 133 & 179 & 151 & 112 \\ \hline \end{tabular} Send data to Excel Part 1 of 6
Following is a boxplot for these data. Is it reasonable to assume that the conditions for performing a hypothesis test are satisfled? Explain.
The boxplot shows that there \square outliers.
The boxplot shows that there \square is no evidence of strong skewness.
We \square assume that the population is approximately normal.
It \square reasonable to assume that the conditions are satisfied.
Part 2 of 6
State the appropriate null and alternate hypotheses. H0:μ=150H1:μ<150\begin{array}{l} H_{0}: \mu=150 \\ H_{1}: \mu<150 \end{array}  This hypothesis test is a left-tailed  test. \text { This hypothesis test is a left-tailed } \quad \nabla \text { test. }
Part: 2/62 / 6
Part 3 of 6
Compute the value of the test statistic. Round the answer to three decimal places. t=1.801t=-1.801 \square ×\times 5 ×\times 5 \square \infty (2) 目 4) \square alo 탕 I \square 0012014016018010\begin{array}{l} 00 \quad 120 \quad 140 \quad 160 \quad 180 \\ 10 \end{array} \qquad

Studdy Solution
Compare the PP-value with α\alpha to make a conclusion.
- If PP-value < α\alpha, reject H0H_0. - If PP-value α\geq \alpha, fail to reject H0H_0.
The test statistic is t=1.801t = -1.801. The PP-value needs to be compared with α=0.01\alpha = 0.01 to conclude whether the mean number of diners decreased.

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