91影视

We have been given that we collected data on the number of shoppers at a random sample of$45$ grocery stores on Friday the $6th$and Friday the $13th$in the same month and the mean difference is $-46.5$and the standard deviation of the differences is $178.0$.

We need to find out that if the result of this study is statistically significant, can we conclude that the difference in shopping behavior is due to the effect of Friday the $13th$on people鈥檚 behavior.

We note that the selected stores are a random sample and we took two Fridays in the same month and it may be possible that some special occasion occurs in either of these two weeks which affects the no. of shoppers on these days.

So we can not conclude that the difference in shopping behavior is due to the effect of Friday the $13th$on people鈥檚 behavior.

We have been given that we collected data on the number of shoppers at a random sample of $45$grocery stores on Friday the $6th$and Friday the $13th$in the same month and the mean difference is $-46.5$and the standard deviation of the differences is $178.0$.

We need to find out that do these data provide convincing evidence at the $伪=0.05$level that the number of shoppers at grocery stores on these two days differs, on average.

Given:

$$\begin{gathered} n=Samplesize=45 \\ {\stackrel{炉}{x}}_d=Meandifferences=-46.5 \\ {s}_d=S\tan darddeviationofdifferences=178 \\ 伪=Significancelevel=0.05 \end{gathered}$$

Now we will carry out a hypothesis test for the population mean difference:

$$\begin{gathered} H_0=Nullhypothesis \\ (H{)}_a=Alternativehypothesis \\ {H}_0:{渭}_d=0 \\ {H}_a:{渭}_d鈮�0 \\ Nowwewilldeterminethevalueofteststatistics: \\ t=\frac{\stackrel{炉}{x_d}-{渭}_d}{{\displaystyle \raisebox{1ex}{$s_d$}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}} \\ t=\frac{-46.5-0}{{\displaystyle \raisebox{1ex}{$178$}\!\left/ \!\raisebox{-1ex}{$\sqrt{45}$}\right.}}鈮�-1.752 \end{gathered}$$

The P-value is the probability of obtaining the value of test statistics.

Degree of freedom$=n-1=45-1=44$

The test is a two-tailed test so we double the boundaries of the P-value and we will use the value of the degree of freedom equal to $40$because the table does not contain value $44$.

$$\begin{gathered} 0.05=2\left(0.025\right)<P<2\left(0.05\right)=0.10 \\ nowatt=-1.752andDegreeoffreedom=40,pvalue=0.08674 \end{gathered}$$

We have to reject the null hypothesis if the probability value is less than the hypothesis value.

$P>0.05鈬�Failtoreject{H}_0$

There is no convincing evidence that the number of shoppers at grocery stores on these two days differs, on average.

We have been given that we collected data on the number of shoppers at a random sample of $45$grocery stores on Friday the $6th$and Friday the $13th$in the same month and the mean difference is $-46.5$and the standard deviation of the differences is $178.0$.

We need to find out that based on part (a) which type of error鈥攁 Type I error or a Type II error鈥攃ould we have made.

We could have Type II error (if we made an error) because we failed to reject the null hypothesis.