Question

.1. Does an average pack of banana bread cookies weighs 40 grams? A random sample of 35 boxes showed a mean of 41.95 grams. The company has specified the population standard deviation to be 5 grams. Test at the 0.10 level of significance.

2. An Insurance Company reports that the mean cost to process claim is $50. An industry comparison showed this amount to be larger than most other insurance companies, so they instituted cost-cutting measures. To evaluate the effect of cost-cutting measure, the company selected a random sample of claims processed last month and determine the cost to process these selected claims. The sample information is reported as follows: 45, 48, 52, 47, 39, 42. At the 0.10 significance level is it reasonable to conclude that the mean cost to produce a claim is now less than $50?

3. A marketing company claims that it receives more than 6.5% responses 6 points from its Mailing. A sample of 600 mails yesterday revealed 45 responses. At the 0.01 level of significance, can we conclude that the claim is true? 

Answer 1

Q1)

Null hypothesis = Estimated mean = 40

Test statistic = (estimated value - mean)/standard error

Standard error = standard deviation/n^0.5

= 5/(35^0.5) = 0.845

Thus, test statistic = (41.95 - 40)/0.845 = 2.307

Z-critical value at 10% leval of significance = 1.645

Since test statistic > z-critical value, the null hypothesis can be rejected and thus the average pack of banana cookie does not weigh 40 grams

Answer 2

1. In this question , we are given with $\overline{X} = 41.95$ , n = 35 , $\sigma = 5$

Step 1 - Set null and alteranate hypothesis as -

$H_{0}:\mu = 40$

$H_{1}:\mu \neq 40$

Step 2 - Use formula of Z statistic '

$Z = \frac{\overline{X}- \mu }{\sigma /\sqrt{n}}$

Step 3 - Substitute values in the formula

$Z = \frac{41.95- 40}{5/\sqrt{35}}$

Step 4 - Compare caculated value from tabulated value of Z at 0.10 level of significance

Z value at 0.10 is 1.28

So,

that means , we reject null hypothesis

Step 5 - Draw conclusion

As Z statistic value is high as compared to value of Z at 0.10 level of significanace. Therefore, we reject null hypothesis and accept alterante hypothesis.

Therefore, we can say that average weight of banana bread cookies does not weigh 40grams.