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 8.5 days. The population is normally distributed with a population standard deviation of 0.9 days. At α=0.02, what type of test is this and can you support the company’s claim using the p-value?

Answer 1

Answer with explanation:

Let $\mu$ be the population mean.

Null hypothesis : $H_0:\mu=8.4$

Alternative hypothesis : $H_1:\mu<8.4$

Since the alternative hypothesis is left tailed, so the test is a left-tailed test.

Sample size : n=106 >30 , so we use z-test.

Test statistic: $z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

$z=\dfrac{8.5-8.4}{\dfrac{0.9}{\sqrt{106}}}=1.14395890455\approx1.14$

The P-value (Left tailed test)= $P(z<1.14)= 0.8728568$

Since, the p-value is greater than the significance level , so we fail to reject the null hypothesis.

Hence, we cannot support the company's claim.