Answer to Question #269462 in Statistics and Probability for Brad

Question

The south African department of health is concerned that the average percentage mark up on a particular vitamin products during dawn of covid 19 is 41% or above. To test their concern, they sampled 40 stores that sell the product, and found that the mean mean percentage mark up was 49.1% assume that the population standard deviation of the percentage mark up is 18.7%

Answer 1

The hypotheses to be tested are,

$H_0:\mu=41$

$Against$

$H_1:\mu\gt41$

The data above gives the values of the population standard deviation, the sample mean and sample size as follows,

$\sigma=18.7,\space \bar{x}=49.1,\space n=40$

Since we know the population standard deviation and the sample size is greater than 30, we shall apply the standard normal distribution to test the department's claim.

The test statistic is given as,

$Z^*=(\bar{x}-\mu)/(\sigma/\sqrt{n})=(49.1-41)/(18.7/\sqrt{40})= 2.739513$

$Z^*$ is compared with the table value at $\alpha=5\%=0.05$ given as, $Z_{0.05}=1.645$ and the null hypothesis is rejected if $Z^*\gt Z_{0.05}$.

Since $Z^*=2.739513\gt Z_{0.05}=1.645$, we reject the null hypothesis and conclude that the  average percentage mark up on a particular vitamin products during dawn of Covid-19 is 41% or above at 5% level of significance.