Answer to Question #343315 in Statistics and Probability for Jhaycee

Question

A new drug on the market is claimed by its manufacturers to reduce overweight women by 4.55 kg per month with standard deviation of 0.91 kg. Ten women chosen at random have reported losing an average of 4.05 kg within a month. Does this data support the claim of the manufacturer at 0.05 level of significance?

Answer 1

Step 1:

Ho: The average weight loss per month using new drug is equal to 4.55 kg.

Ha: The average weight loss per month using new drug is not equal to 4.55 kg.

$H_0:\mu=4.55$

$H_a:\mu\not=4.55$

Step 2: Type of test: Two- tailed test or non- directional test.

Based on the information provided, the significance level is $\alpha=0.05,$ and the critical value for a two-tailed test is $z_c=1.96.$ We found it using t-distribution table:

The rejection region for this two-tailed test is $R=\{z:|z|>1.96\}.$

Step 3: Compute the test value using the formula or test statistic.

Given: $X=4.05, \mu=4.55, n=10, \sigma=0.91$

Substitute the given values in the formula:

$Z= \frac {(X-\mu)\sqrt{n}}{\sigma}=\frac {(4.05-4.55)\sqrt{10}}{0.91}=​-1.74$

Step 4: Decision: Since the computed or test value does not fall within the rejection region , we accept the null hypothesis.

Step 5: Conclusion: There is no significant difference between the sample mean and population mean. Thus, the manufacturer is correct in claiming that the new drug in the market can reduce overweight women by 4.55 kg per month.