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Running head: STATISTICS RESEARCH PROJECT INSTRUCTIONS 1
Statistics Research Project Instructions
Ti’Wan Gamble
MATH 201-D05 LUO
Liberty University
8/10/2020
STATISTICS RESEARCH PROJECT INSTRUCTIONS 2
Statistics Research Project Instructions
PART 4 – HYPOTHESIS TESTING WEEK 6
Based on your sample, you will conduct a hypothesis test with 𝛼 = 0.05 to test two of the claims
of the above article. Using the same sheet as last week, answer the following in the β€œweek 6” tab:
1) Claim: the average age of online students is 32 years old.
a. What is the null hypothesis?
b. What is the alternative hypothesis?
c. What distribution should be used?
d. What is the test statistic?
e. What is the p-value?
f. What is the conclusion?
Claim: the average age of online students is 32 years old
a) Null Hypothesis: The population mean of online students is equal to 32 years old
b) Alternative Hypothesis: The population mean of online students is not equal to 32 years
old
c) A T distribution should be used.
d) Test statistic
𝑑 π‘ π‘‘π‘Žπ‘‘π‘–π‘ π‘‘π‘–π‘ = (π‘₯ξͺ§ βˆ’Β΅)
𝑠
βˆšπ‘›
Where π‘₯ξͺ§ = 31.294, s = 10.178, Β΅ = 32, n = 68
𝑑𝑒𝑠𝑑 π‘ π‘‘π‘Žπ‘‘π‘–π‘ π‘‘π‘–π‘ = (31.294βˆ’32)
10.178
√68 =βˆ’0.572
Test statistics: - 0.57
e) P - Value
From the t distribution the p –value = 0.5706
f) Conclusion
Fail to reject the null hypothesis. There is sufficient evidence to suggest that the average age of
online students is 32 years old
STATISTICS RESEARCH PROJECT INSTRUCTIONS 3
ο‚· 2) Claim: the proportion of males in online classes is 35%
a) What is the null hypothesis?
b) What is the alternative hypothesis?
c) What distribution should be used?
d) What is the test statistic?
e) What is the p-value?
f) What is the conclusion?
Claim: The proportion of males in online classes is 35%
a) Null Hypothesis: The population proportion of males in online classes is equal to 35%
b) Alternative Hypothesis: The population proportion of males in online classes is not equal
to 35%
c) A Normal distribution
d)
𝑑𝑒𝑠𝑑 π‘ π‘‘π‘Žπ‘‘π‘–π‘ π‘‘π‘–π‘ = (π‘ξžΈ βˆ’π‘)
βˆšπ‘(1βˆ’π‘)
𝑛
Where π‘ξžΈ = sample proportion, 𝑝 = population proportion under the null hypothesis
π‘ξžΈ = 0.4559 𝑝 =0.35
𝑑𝑒𝑠𝑑 π‘ π‘‘π‘Žπ‘‘π‘–π‘ π‘‘π‘–π‘π‘  = (0.4559βˆ’0.35)
√0.35(1βˆ’0.35)
68 =1.8309
Test Statistics: 1.83
e) P-value
P – Value = 2*P (Z > 1.83)
From the standard normal tables, P (Z > 1.83) = 0.0336
STATISTICS RESEARCH PROJECT INSTRUCTIONS 4
P – Value = 2*0.0336 =0.0672
P – Value =0.0672
f) Conclusion
Do not reject the null hypothesis. There is sufficient evidence to conclude that the
proportion of males in online classes is 35%
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