A manufacturer of a new type of fishing line claims that the mean breaking stren

Question

A manufacturer of a new type of fishing line claims that the mean breaking strength of the line is 15 kilograms. To test this claim, a random sample of 50 pieces of line are tested and found to have a sample average breaking strength of 15.3 kilograms with a sample standard deviation 8 kilograms. Does this data support the manufacturer's claim? State the null hypothesis. State the alternative hypothesis? Compute the P-value of the test and state the conclusion you would draw assuming a significance level of 05. A public utility company wishes to determine whether its new work schedule has significantly reduced the waiting time for customers which had previously averaged 30 minutes. A random sample of 25 customer waiting times yielded a mean of 28 minutes and a sample standard deviation of 6 minutes. Assuming customer waiting time is approximately normally distributed, does this data indicate a reduction in customer waiting time? State the null hypothesis. State the alternative hypothesis. Compute the P-value for the test and draw a conclusion assuming a 05 significance level.

Explanation / Answer

Q.4 (a) Null HYpothesis is that the mean breaking strength of the fishing line is 15 kilograms. = 15 kg

(b) ALternaive Hypothesis is that the mean breaking strenght of the fishing line is not equal to 15 kilograms. 15 kg.

(c) P - value will be calculated by using t - test as sample standard deviation is given.

Here sample mean xbar = 15.3

standard deviation of the sample s = 0.8 kg

t = (xbar - H)/ (s/n) = (15.3 - 15)/ (0.8/ 50) = 0.3/ 0.1131 = 2.6525

so P - value = 2* Pr (xbar > 15.3 ; 15; 0.8/50) = 0.0107 < 0.05

so we can reject the null hypothesis and can conclude that the mean strength of fishing line is not equal to 15 kg.

Q.5

(a) Null HYpothesis is that new work schedule has not reduced waiting time for customers. t = 30 minutes

(b) Alternaive Hypothesis is that the new work schedule has significantly reduced waiting time for customers. t < 30 minutes

(c) P - value will be calculated by using t - test as sample standard deviation is given.

Here sample mean xbar = 28 minutes

standard deviation of the sample s = 6 minutes

t = (xbar - H)/ (s/n) = (28 - 30)/ (6/ 25) = -2/ 1.2 = -1.67

so P - value = Pr (xbar < 28 ; 30; 6/25) = 0.0543 > 0.05

so we cannot reject the null hypothesis and can conlude that new work schedule has not reduced waiting time for customers.

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