Question

A pharmaceutical company claims that the average cold lasts an average of 8.4 days. They are using this as a basis to test new medicines designed to shorten the length of colds. A random sample of 106 people with colds, finds that on average their colds last 8.7 days. The population is normally distributed with a standard deviation of 0.9 days. At a-0.02, what type of test is this and can you support the company's claim using the p-value? a. Claim is null, reject the null and cannot support claim as the p-value (0.001) is less than alpha (0.02) b. Claim is null, fail to reject the null and support claim as the p-value (0.001) is greater than alpha (0.02) c. Claim is alternative, fail to reject the null and cannot support claim as the p-value (0.000) is less than alpha (0.02) d. Claim is alternative, reject the null and support claim as the p-value (0.000) is greater than alpha (0.02)

Answer 1

`z=(8.7-8.4)/((0.9)/(√(106)))=3.432`

Since is a one side right tailed test the p value would be:

`p_v =P(Z>3.432)=0.0003`

If we compare the p value and a significance level given
`\alpha=0.02` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis.

And the best conclusion would be:

a. Claim is null, reject the null and cannot support claim as the p-value (0.001) is less than alpha (0.02)

Explanation:

Data given and notation

`\bar X=8.7` represent the average for the sample

`\sigma=0.9` represent the population standard deviation

`n=106` sample size

`\mu_o =8.4` represent the value that we want to test

`\alpha=0.02` represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to apply a one upper tailed test.

What are H0 and Ha for this study?

Null hypothesis:
`\mu \geq 8.4`

Alternative hypothesis :
`\mu > 8.4`

Compute the test statistic

The statistic for this case is given by:

`z=(\bar X-\mu_o)/((\sigma)/(√(n)))` (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

`z=(8.7-8.4)/((0.9)/(√(106)))=3.432`

Critical value

On this case since we have a right tailed test we need to look into the normal standard distribution and find a value that accumulates 0.02 of the area on the right of the distribution or 0.98 of the area on the left and for this case we see that
`z_(critical)=2.054`

Give the appropriate conclusion for the test

Since is a one side right tailed test the p value would be:

`p_v =P(Z>3.432)=0.0003`

Conclusion

If we compare the p value and a significance level given
`\alpha=0.02` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis.

And the best conclusion would be:

a. Claim is null, reject the null and cannot support claim as the p-value (0.001) is less than alpha (0.02)