Question

A) The production supervisor of an assembly line for producing potato chips suspects that the weight of bags of potato chips produced is higher than the specification of 250 mg. She would like to take a simple random sample of bags of potato chips produced by the assembly line and perform a statistic test to confirm that this is indeed the case. (6 points)

1. What would be the null and alternative hypothesis for this research?
2. What is the type I error in this situation?
3. What is the type II error in this situation?                                      

B) A simple random sample of 40 accounts is taken from an account receivables portfolio of ABC Ltd and the average account balance is $750. The population standard deviation σ is known to be $80. (6 points)

1. Test the hypothesis that the population mean account balance is greater than $680 using the p-value approach and a 0.05 level of significance.
2. Test the hypothesis that the population mean account balance is less than $820 using the critical value approach and a 0.05 level of significance.
3. Test the hypothesis that the population mean account balance is different from $800. using the p-value approach and a 0.05 level of significance

C) A simple random sample of 80 customers is taken from a customer information file and the average age is 34. The population standard deviation σ is unknown. Instead the sample standard deviation s is also calculated from the sample and is found to be 5.5. (6 points)

1. Test the hypothesis that the population mean age is greater than 30 using the critical value approach and a 0.05 level of significance.
2. Test the hypothesis that the population mean age is less than 37 using the p-value approach and a 0.05 level of significance.
3. Test the hypothesis that the population mean age is different from 31 using the p-value approach and a 0.05 level of significance.

D) The proportion of customers who are completely satisfied in a recent satisfaction survey of 400 customers at XYC Inc. is found to be 0.24. (6 points)

1. Test the hypothesis that the population proportion of customers who are completely satisfied is greater than 0.21 using the critical value approach and a 0.05 level of significance.
2. Test the hypothesis that the population proportion of customers who are completely satisfied is less than 0.28 using the p-value approach and a 0.05 level of significance.
3. Test the hypothesis that the population proportion of customers who are completely satisfied is different from 0.22 using the p-value approach and a 0.05 level of significance.

Answer 1

A)

1: Hypotheses are:

Ho : μ = 250

$H_{a}:\mu>250$

2:

Type I error is the probability of rejecting the true null hypothesis.

Researcher incorrectly conclude that the weight of bags of potato chips produced is higher than the specification of 250 mg while actually it is 250mg.

3:

Type II error is the probability of fail to reject the false null hypothesis.

Researcher incorrectly conclude that the weight of bags of potato chips produced is 250 mg while actually it is higher than 250mg.

b)(1)

Here we have -750,c-80, n-40 X Hypotheses are: H 0: μ <=680 Level of significance: α =0.05 Test is one tailed (right tailed). Test Statistics: (750-680)/(80/sqrt(40)) σ/Vi, = 5.53 Critical value: 1.645 Rejection Region Ifz>1.645, Reject HO Decision: Since test statistics lies in rejection region so we reject the null hypothesis. P-value: P-value-0 Decision: Since p-value is less than level of significance so we reject the null hypothesis. Excel function for critical value: NORMSINV(1-0.05) Excel function for p-value: -1-NORMSDIST(5.53)

(2)

Here we have -750, σ-80, n-40 x Hypotheses are Ho 820 Ha:μ< 820 Level of significance: α-0.05 Test is one tailed (left tailed). Test Statistics: (750-820)/(80/sqrt(40) --5.53 Critical value: 1.645 If z<-1.645, Reject HO Rejection Region: Decision: Since test statistics lies in rejection region so we reject the null hypothesis P-value: P-value-O0 Decision: Since p-value is less than level of significance so we reject the nul hypothesis Excel function for critical value: NORMSINV(0.05) Excel function for p-value: NORMSDIST(-5.53)

(3)

Here we have x:750,0-80, n -40 Hypotheses are: Ha: μ# 800 Level of significance: 0.05 α Test is two tailed. Test Statistics: x-μ alm = (750-800)/(80/sqrt(40)) -3.95 14 Critical value: If I Rejection Region: > 1.96, Reject H0 Decision: Since test statistics lies in rejection region so we reject the null hypothesis p-value: P-value-0.0001 Decision: Since p-value is less than level of significance so we reject the null hypothesis Excel function for critical value:NORMSINV(1-0.05) -2*(1-NORMSDIST(3.95)) Excel function for p-value: