However, a random sample of 30 US women has a mean height of 67 inches with standard deviation of 6.2 inches. At 0.05 level of significance, determine if the data indicates that women mean height is more than 65 inches. Step 1. Write Ho and Ha Ho: μ = 65 Ha: μ> 65 (one-tailed) (we can reject Ho by showing that u is greater than specified in μ Ho) Step 2. Given sample mean is part of a sample mean distribution (sampling X-distribution). Assuming Ho is true, find the mean (μ) and standard deviation (0) of sample mean distribution. sample size = n = 30 Assume Ho is true: μ = 65, then mean of sample mean distribution = μ₁x = μ = 65 std. dev of sample mean distribution == σ Step 3. Model the sample mean distribution as t-graph and find the t-value (t) of the given sample mean. 6.2 Vn Vn √30 given sample mean= x = 67 t₁ = x-μx 67-65 ox Step 4. Find t-critical value (ter) degree of freedom = df =n -1 =30-1=29 level of significance: tail-area: a = 0.05 number of tails occupying the above area: ntails = one ter = ter(df, a, ntails) = ter(29, 0.05, one) =1.70 (from table) 1.132 1.132 yes. Reject Ho = 1.132 = 1.77 Step 5. Compare t* and ter and determine if t* larger than ter ignoring signs Is t*> ter (ignoring signs)? Is 1.77 > 1.70?
However, a random sample of 30 US women has a mean height of 67 inches with standard deviation of 6.2 inches. At 0.05 level of significance, determine if the data indicates that women mean height is more than 65 inches. Step 1. Write Ho and Ha Ho: μ = 65 Ha: μ> 65 (one-tailed) (we can reject Ho by showing that u is greater than specified in μ Ho) Step 2. Given sample mean is part of a sample mean distribution (sampling X-distribution). Assuming Ho is true, find the mean (μ) and standard deviation (0) of sample mean distribution. sample size = n = 30 Assume Ho is true: μ = 65, then mean of sample mean distribution = μ₁x = μ = 65 std. dev of sample mean distribution == σ Step 3. Model the sample mean distribution as t-graph and find the t-value (t) of the given sample mean. 6.2 Vn Vn √30 given sample mean= x = 67 t₁ = x-μx 67-65 ox Step 4. Find t-critical value (ter) degree of freedom = df =n -1 =30-1=29 level of significance: tail-area: a = 0.05 number of tails occupying the above area: ntails = one ter = ter(df, a, ntails) = ter(29, 0.05, one) =1.70 (from table) 1.132 1.132 yes. Reject Ho = 1.132 = 1.77 Step 5. Compare t* and ter and determine if t* larger than ter ignoring signs Is t*> ter (ignoring signs)? Is 1.77 > 1.70?
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section: Chapter Questions
Problem 22SGR
Related questions
Question
How can I write a concluding statement about Ho for this problem?
Transcribed Image Text:The mean height of a US woman is believed to be at most 65 inches (65 or less).
However, a random sample of 30 US women has a mean height of 67 inches with
standard deviation of 6.2 inches. At 0.05 level of significance, determine if the data
indicates that women mean height is more than 65 inches.
Step 1. Write Ho and Ha
Ho: μ = 65
Ha: μ> 65 (one-tailed) (we can reject Ho by showing that μ is greater than specified in
Ho)
Step 2. Given sample mean is part of a sample mean distribution (sampling
x-distribution). Assuming Ho is true, find the mean (u) and standard deviation (0)
of sample mean distribution.
sample size = n = 30
Assume Ho is true: μ = 65, then
mean of sample mean distribution = μx = μ = 65
std. dev of sample mean distribution = 0x
0
S
6.2
√n √n √30
Step 3. Model the sample mean distribution as t-graph and find the t-value (t) of
the given sample mean.
given sample mean= x = 67
x-μx 67-65
t₁ =
2
1.132
ox
1.132
= 1.132
= 1.77
Step 4. Find t-critical value (ter)
degree of freedom = df=n -1 = 30 - 1 = 29
level of significance: tail-area: a = 0.05
number of tails occupying the above area: ntails = one
ter ter(df, a, ntails) = ter(29, 0.05, one) =1.70 (from table)
yes.
Reject Ho
Step 5. Compare t* and ter and determine if t* larger than ter ignoring signs
Is t > ter (ignoring signs)?
Is 1.77 > 1.70?
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