Question

CCN and ActMedia provided a television channel targeted to individuals waiting in supermarket checkout lines. The channel showed news, short features, and advertisements. The length of the program was based on the assumption that the population mean time a shopper stands in a supermarket checkout line is 8 minutes. A sample of actual waiting times will be used to test this assumption and determine whether actual mean waiting time differs from the standard.a. Formulate the hypothesis for this application.b. A sample of 120 shoppers showed a sample mean waiting time of 8.5 minutes. Assume a population standard deviation of 3.2 minuest. What is the p-value?c. At a=.0t what is your conclusion?d. Compute a 95% confidence interval for the population mean. Does it support your conclusion?

Answer 1

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ = 8 minutes    

    Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 8 minutes

(b) The P-value of the test statistics is 0.0436.

(c) We conclude that the actual mean waiting time equals the standard at 0.05 significance level.

(d) 95% confidence interval for the population mean is [7.93 minutes , 9.07 minutes].

Step-by-step explanation:

We are given that the length of the program was based on the assumption that the population mean time a shopper stands in a supermarket checkout line is 8 minutes.

A sample of 120 shoppers showed a sample mean waiting time of 8.5 minutes. Assume a population standard deviation of 3.2 minutes.

Let $\mu$ = actual mean waiting time.

(a) So, Null Hypothesis, $H_0$ : $\mu$ = 8 minutes     {means that the actual mean waiting time does not differs from the standard}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 8 minutes     {means that the actual mean waiting time differs from the standard}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

                        T.S. =  $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$  ~ N(0,1)

where, $\bar X$ = sample mean waiting time = 8.5 minutes

            $\sigma$ = population standard deviation = 3.2 minutes

            n = sample of shoppers = 120

So, test statistics  =  $\frac{8.5-8}{\frac{3.2}{\sqrt{120} } }$  

                               =  1.71

The value of t test statistics is 1.71.

(b) Now, the P-value of the test statistics is given by;

                    P-value = P(Z > 1.71) = 1 - P(Z $\leq$ 1.71)

                                  = 1 - 0.9564 = 0.0436

(c) Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the actual mean waiting time equals the standard.

(d) Now, the pivotal quantity for 95% confidence interval for the population mean is given by;

                        P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$  ~ N(0,1)

where, $\bar X$ = sample mean waiting time = 8.5 minutes

            $\sigma$ = population standard deviation = 3.2 minutes

            n = sample of shoppers = 120

            $\mu$ = population mean

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the population proportion, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                             of significance are -1.96 & 1.96}  

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times{\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times{\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times{\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times{\frac{\sigma}{\sqrt{n} } }$ ]

  = [ $8.5-1.96 \times{\frac{3.2}{\sqrt{120} } }$ , $8.5+1.96 \times{\frac{3.2}{\sqrt{120} } }$ ]

  = [7.93 minutes , 9.07 minutes]

Therefore, 95% confidence interval for the population mean is [7.93 minutes , 9.07 minutes].

Yes, it support our above conclusion.