Question

The mean waiting time at the​ drive-through of a​ fast-food restaurant from the time an order is placed to the time the order is received is 86.3 seconds. A manager devises a new​ drive-through system that she believes will decrease wait time. As a​ test, she initiates the new system at her restaurant and measures the wait time for 10 randomly selected orders. The wait times are provided in the table to the right. Complete parts​ (a) and​ (b) below. 105.5 79.4 68.0 94.2 59.1 86.1 76.7 72.2 65.7 83.1 LOADING... Click the icon to view the table of correlation coefficient critical values. ​(a) Because the sample size is​ small, the manager must verify that the wait time is normally distributed and the sample does not contain any outliers. The normal probability plot is shown below and the sample correlation coefficient is known to be requals0.989. Are the conditions for testing the hypothesis​ satisfied? ▼ Yes, No, the conditions ▼ are not are satisfied. The normal probability plot ▼ is not is linear​ enough, since the correlation coefficient is ▼ less greater than the critical value. 60 75 90 105 -2 -1 0 1 2 Time (sec) Expected z-score A normal probability plot has a horizontal axis labeled Time (seconds) from 50 to 115 in increments of 5 and a vertical axis labeled Expected z-score from negative 2 to 2 in increments of 0.5. Ten plotted points closely follow the pattern of a line that rises from left to right through (59, negative 1.55) and (94, 1). All coordinates are approximate. ​(b) Is the new system​ effective? Conduct a hypothesis test using the​ P-value approach and a level of significance of alpha equals 0.05. First determine the appropriate hypotheses. Upper H 0​: ▼ mu sigma p ▼ less than not equals equals greater than 86.3 Upper H 1​: ▼ mu sigma p ▼ greater than equals not equals less than 86.3 Find the test statistic. t 0equals nothing ​(Round to two decimal places as​ needed.) Find the​ P-value. The​ P-value is nothing. ​(Round to three decimal places as​ needed.) Use the alpha equals 0.05 level of significance. What can be concluded from the hypothesis​ test? A. The​ P-value is less than the level of significance so there isnbsp not nbspsufficient evidence to conclude the new system is effective. B. The​ P-value is greater than the level of significance so there isnbsp not nbspsufficient evidence to conclude the new system is effective. C. The​ P-value is less than the level of significance so there isnbspsufficient evidence to conclude the new system is effective. D. The​ P-value is greater than the level of significance so there isnbspsufficient evidence to conclude the new system is effective. Click to select your answer(s).

Answer 1

Mean	79
Standard Deviation (s)	13.92879
Skewness	0.54775
Kurtosis	0.03434
Lowest Score	59.1
Highest Score	105.5
Distribution Range	46.4
Total Number of Scores	10
Number of Distinct Scores	10
Lowest Class Value	50
Highest Class Value	105.9
Number of Classes	4
Class Range	14

a) Yes, conditions are satisfied.

b)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: μ > 86.3

Alternative hypothesis: μ < 86.3

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 4.40464

DF = n - 1 = 10 - 1

D.F = 9

t = (x - μ) / SE

t = -1.66

where s is the standard deviation of the sample, x is the sample mean, μ is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of -1.66. We use the t Distribution Calculator to find P(t > -1.66) = 0.06565.

Thus the P-value in this analysis is 0.06565.

Interpret results. Since the P-value (0.06565) is greater than the significance level (0.05), we cannot reject the null hypothesis.

P-value is greater than the level of significance so there is sufficient evidence to conclude the new system is effective.