A mail-order business prides itself in its ability to fill customers\' orders in

Question

A mail-order business prides itself in its ability to fill customers' orders in less than six calendar days, on average. Periodically, the operations manager selects a random sample of customer orders and determines the number of days required to fill the orders. On one occasion when a sample of 39 orders was selected, the average number of days was 6.65 with a population standard deviation of 1.5 days. Apply the p-value and the critical value approaches to test the hypotheses at a significance level alpha = .10. The manufacturer of an over-the-counter heartburn relief mediation claims that it product brings relief in less than 3.5 minutes, on average. To be able to make this claim the manufacturer was required by the FDA to present statistical evidence in support of the claim. The manufacturer reported that for a sample of 50 heartburn sufferers, the mean time to relief was 3.3 minutes and the population standard deviation was 1.1 minutes. Test the hypotheses at a significance level alpha = .05 using the critical value approach.

Explanation / Answer

Solution:-

Given, Mean = 6 calender days

Sample size choosen = 39

Computed mean = 6.65 with standard deviation of 1.5

Significance level, alpha = 0.10

The solution to this problem takes four steps:

(1) state the hypotheses,

(2) formulate an analysis plan,

(3) analyze sample data, and

(4) interpret results.

We work through these steps below:

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

Null hypothesis: = 6
Alternative hypothesis: 6

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

Formulate an analysis plan. For this analysis, the significance level is 0.10. 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) = 1.5 / sqrt(39) = 0.24019
DF = n - 1 = 39 - 1 = 38
t = (x - ) / SE = (6.65 - 6)/0.24019 = 2.70619

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

Since we have a two-tailed test, the P-value is the probability that the t statistic having 38 degrees of freedom is less than 2.70619 or greater than 2.70619.

We use the t Distribution Calculator to find P(t < 2.70619)

The P-Value is 0.010138.

The result is significant at p < 0.10.

Interpret results. Since the P-value (0.010138) is smaller than the significance level (0.10), we can reject the null hypothesis.That is accepting the alternate hypothesis where, mean was not equal to 6, that mean it takes more than 6 days to deliver.

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