Answer to Question #222097 in Statistics and Probability for Darryl

Question

a) The Ghana Economic Management Association wishes to have information on the mean income of store managers in the retail industry. A random sample of 256 managers reveals a sample mean of Ghc45,420. The standard deviation of this population is Ghc2,050. What is a reasonable range of values for the population mean? Interpret your results. 

b) i. Briefly outline the steps involved in hypothesis testing 

ii. Use a table to illustrate how Type I and Type II error can occur

Answer 1

Part a

90% CI for $\mu$ using Normal dist

Sample Mean $| \bar{x} = 45420$

Since the sample size is large, sample standard deviation can be used to approximate population standard deviation.

Population Standard deviation $=|\sigma =2050$

Sample Size $=| n = 256$

Significance level $= \alpha =1— Confidence =1-0.9 = 0.1$

The Critical Value$= z_{\alpha/2} =z_{0.06} =1.645$ (From z table , using interpolation $\frac{1}{2}$ the distance between 164 and 1.65 )

Critical Values , $± z_{\alpha/ 2}=±1.645$

Margin of Error $= E = z_{\alpha/2}*\frac{\sigma}{\sqrt{n}} =1.645 *\frac{2050}{\sqrt{256}}=210.765625$

Margin of Error, $E= 210.766$

Limits of 90% confidence interval are given by

Lower limit $= \bar{x} - E = 45420 — 210.365625 = 45209.234375 = 45 209.234$

Upper limit $=\bar{x}+ E = 45420 +210.765625 = 45630.365625 = 45 630.766$  

90% confidence interval is $\bar{x}±E = 45420± 210.765625 = (45209.234375, 45630.765625)$

90% CI using normal dist: $(45 202.234) < \mu < (45 630.766)$

95% CI for $\mu$ using Normal dist

Sample Mean $| \bar{x} = 45420$

Since the sample size is large, sample standard deviation can be used to approximate population standard deviation.

Population Standard deviation $=|\sigma =2050$

Sample Size $=| n = 256$

Significance level $= \alpha =1— Confidence =1-0.95 = 0.05$

The Critical Value$= z_{\alpha/2} =z_{0.025} =1.96$

Critical Values , $± z_{\alpha/ 2}=±1.96$

Margin of Error $= E = z_{\alpha/2}*\frac{\sigma}{\sqrt{n}} =1.96 *\frac{2050}{\sqrt{256}}=251.125$

Margin of Error, $E= 251.125$

Limits of 95% confidence interval are given by

Lower limit $= \bar{x} - E = 45420 - 251.125 = 45168.875$

Upper limit $=\bar{x}+ E = 45420 +251.125 = 45671.125$  

95% confidence interval is $\bar{x}±E = 45420± 251.125 = (45168.875,45671.125)$

90% CI using normal dist: $(45168.875) < \mu < (45671.125)$

99% CI for $\mu$ using Normal dist

Sample Mean $| \bar{x} = 45420$

Population Standard deviation $=|\sigma =2050$

Sample Size $=| n = 256$

Significance level $= \alpha =1— Confidence =1-0.99 = 0.01$

The Critical Value$= z_{\alpha/2} =z_{0.005} =2.58$

Critical Values , $± z_{\alpha/ 2}=±2.58$

Margin of Error $= E = z_{\alpha/2}*\frac{\sigma}{\sqrt{n}} =2.58 *\frac{2050}{\sqrt{256}}=330.5625$

Margin of Error, $E= 330.5625$

Limits of 99% confidence interval are given by

Lower limit $= \bar{x} - E = 45420 - 303.563 = 45089.438$

Upper limit $=\bar{x}+ E = 45420 +303.5625= 45750.563$  

99% confidence interval is $\bar{x}±E = 45420± 303.5625 = (45089.438,45750.503)$

99% CI using normal dist: $(45089.438) < \mu < (45750.503)$

The widest CI is the one with highest confidence level. 

99% CI is the widest. 

Part b

i)

Enter the Null Hypothesis.

Describe the Alternative Hypothesis.

Determine the Significance Level (a)

Determine the Test Statistic and the Corresponding P-Value.

Making a decision.

ii)

In statistics, a Type I error means rejecting the null hypothesis when it's actually true. In contrast, a Type II error means failing to reject the null hypothesis when it's actually false.