Question

Each answer requires:

1. The problem statement
2. Assumptions
3. The null and alternate hypothesis statements
4. The significance level
5. The test statistic (as an equation)
6. Decision rules
7. The calculated value of the test statistic
8. The p-value
9. Interpret the results of the test.

1. When operating normally, a manufacturing process produces tablets for which the mean weight of the active ingredient is 5 grams, and the standard deviation is 0.025 gram. For a random sample of 12 tables the following weights of active ingredient (in grams) were found:

5.01 4.69 5.03 4.98 4.98 4.95 5.00 5.00 5.03 5.01 5.04 4.95

1. Without assuming that the population variance is known, test the null hypothesis that the population mean weight of active ingredient per tablet is 5 grams. Use a two-sided alternative and a 5% significance level. State any assumptions that you make.

2. The president of Amalgamated Retailers International, Sam Peterson, has asked for your assistance in studying the market penetration for the company’s new cell phone. You are asked to determine if the market share is equal to the company’s claim of 35%. You obtain a random sample of potential customers from the area. The sample indicates that 258 out of a total sample of 800 indicate they will purchase from Amalgamated

a) Using a probability of error α=0.03, test the hypothesis that the market share equals the company’s claim of 35% versus the hypothesis that the market share is not equal to the company’s claim.

b) Using a probability of error α=0.03, test the hypothesis that the market share equals the company’s claim of 35% versus the hypothesis that the market share is less than the company’s claim.

Answer 1

Q1) a) Assuming standard deviation is equal to 0.025.

we will set up the null hypothesis that

0

1

under the null hypothesis the test statistics is

Z =
Where
g=0.025

and

5.014.69 5.03 4.95 4.9725 12 n i-1

4.9725 5 Z 0.025 /12

Z=3.81

Z-Tabulated at 5% level of significance is equal to 1.96 which is less then -3.81 hence we will reject our null hypothesis and conclude that $\mu \neq5$.

Q1) b) Assuming standard deviation is not equal to 0.025.

we will set up the null hypothesis that

0

1

under the null hypothesis the test statistics is

  

t S /n

Where

2)2 n s S n 1

$S=\sqrt{\frac{(5.01-4.9725)^{2}+(4.69-4.9725)^{2}+...+(4.95-4.9725)^{2}}{12-1}}=0.0936$

4.9725 5 .t = 0.0936 V12

$t=-1.02$

t-Tabulated at 5% level of significance at 11 df is equal to 2.201 which is greater then -1.02 hence we will accept our null hypothesis and conclude that $\mu = 5$.

Q2) a) we will set up the null hypothesis that

$H_{0}:p=0.35$

$H_{1}:p\neq 0.35$

Under the null hypothesis the test statistics is

$Z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

Where $\hat{p}=\frac{258}{800}=0.3225$

$Z=\frac{0.3225-0.35}{\sqrt{\frac{0.35(1-0.35)}{800}}}$

$Z=-1.63$ and corresponding P.Value = 0.103

Since calculated P.Value = 0.103 which is greater then 0.03. hence we will accept null hypothesis and conclude that p=0.35.

b) we will set up the null hypothesis that

$H_{0}:p=0.35$

Hp<0.35

Under the null hypothesis the test statistics is

$Z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

Where $\hat{p}=\frac{258}{800}=0.3225$

$Z=\frac{0.3225-0.35}{\sqrt{\frac{0.35(1-0.35)}{800}}}$

$Z=-1.63$ and corresponding P.Value =0.055

Since calculated P.Value = 0.055 which is greater then 0.03. hence we will accept null hypothesis and conclude that p=0.35.