Question

A manufacturer claims that the calling range (in feet) of its 900-MHz cordless telephone is greater than that of its leading competitor. A sample of 11 phones from the manufacturer had a mean range of 1390 feet with a standard deviation of 33 feet. A sample of 20 similar phones from its competitor had a mean range of 1360 feet with a standard deviation of 30 feet: Do the results support the manufacturer's claim? Let u be the true mean range of the manufacturer's cordless telephone and uz be the true mean range of the competitor's cordless telephone. Use a significance level of a = 0.05 for the test. Assume that the population variances are equal and that the two populations are normally distributed. Step 3 of 4: Determine the decision rule for rejecting the null hypothesis H. Round your answer to three decimal places. Answer Tables Кеурас Keyboard Shortci Previous Step Answe Reject H, if

Answer 1

In hypothesis testing to compare the calling ranges of two manufacturers' cordless telephones, we set up a null hypothesis of no difference and an alternative claim of greater range. At a 0.05 significance level, the decision rule is to reject the null if the p-value is less than 0.05, which is determined by using a t-test and comparing the test statistic.

Step-by-step explanation:

Hypothesis Testing

To determine if the manufacturer's claim that its cordless telephone has a greater range than its competitor's, we perform a hypothesis test. The null hypothesis (H₀) states that there is no difference in means of the calling ranges (u = u₂), while the alternative hypothesis (H₁) supports the manufacturer's claim (u > u₂). At a significance level of α = 0.05, we use a t-test for two independent samples assuming equal variances.

The decision rule for rejecting the null hypothesis involves calculating the test statistic's p-value and comparing it with the significance level (α). If the p-value is less than α, we reject the null hypothesis. For this test, the decision rule can be expressed as: Reject H₀ if p-value < 0.05.

The calculation of the test statistic and p-value requires the use of statistical software or tables. An example of this calculation can be seen when using a t-table, the calculated t-statistic is compared against a critical value. If the t-statistic is greater than the critical value at a 0.05 significance level for the degrees of freedom determined by the sample sizes, H₀ is rejected.