Question

Hypothesis test for the difference of population means: Z test A study conducted by the research department of a pharmaceutical company claims that the annual spending (per person) for prescription drugs for allergy relief, is greater than or equal to the annual spending (per person) for non-prescription allergy relief medicine, Hly. A health insurance company conducted an independent study and collected data from a random sample of 285 individuals for prescription allergy relief medicine. The sample mean is found to be 17.7 dollars/year, with a sample standard deviation of 4.5 dollars/year. They have also collected data for non-prescription allergy relief medicine. An independent random sample of 200 individuals yielded a sample mean of 18.3 dollars/year, and a sample standard deviation of 3.5 dollars/year. Since the sample size is quite large, it is assumed that the population standard deviation of the sales (per person) for prescription and non-prescription allergy relief medicine can be estimated by using the sample standard deviation values given above. Is there sufficient evidence to reject the claim made by the research department of the company, at the 0.1 level of significance? Perform a one-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.) a P The bull hypothesis: x S P The alternative hypothesis: DO OSO OD The type of test statistic: (Choose one) D<D > The value of the test statistics (Round to at least three decimal places.) x ? The critical value at the 0.1 Level of significance: (Round to at least three decimal places) Can we reject the claim that the mean spending on prescription allergy relief medication is greater than or equal to the mean spending on non-prescription allergy relief medication? Yes ONO

Answer 1

Null Hypothesis H0: $\mu_1 = \mu_2$

Altenative Hypothesis H1:   $\mu_1 > \mu_2$

As, we assume the sample standard deviation as estimate of population standard deviation due to large sample size, the type os test statistic is Z.

Standard error of mean, SE = $\sqrt{\sigma^2_1/n_1 + \sigma^2_2/n_2} = \sqrt{4.5^2/285 + 3.5^2/200}$

= 0.3637342871

Test statistic, Z = (x1 - x2) / SE = (17.7 - 18.3) / 0.3637342871 = -1.649555792

Critical value for right tail test at 0.1 level of significance is  1.282

As, the calculated test statistic is less than the critical value, we fail to reject the claim.  No