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7. Questions 4, 5, and 6 all have 90% confidence level. Why are the crit values different for all three?

7. Questions 4, 5, and 6 all have 90% confidence level. Why are the crit values different for all three?

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.6: Summarizing Categorical Data
Problem 26PPS
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Only Question 7 i need the answer 

 
 
 
7:36 PM Fri Nov 17 Tt O 4. It has been accepted that 40% or less of households changed their driving habits hat 40% because of escalating gas prices. However, a popular energy company challenges that % less than 40% of households have changed their driving habits because of escalating gas prices. 28. of 180 at 70 hous Their recent survey of 180 households finds that 70 households have made lifestyle changes due to escalating gas prices. (hint: first calculate the sample proportion) At a 10% level of significance, test the energy company's challenge. Ho: 0.40 Ha: p<0.40 α = 0.10 po= 40% n=180 X: 70 -1.28 0 p= X = 2080² 器 Z= 0.3889-0.40 0.3889x(1-03889) 180 P = P(Z < - 1:39) + ( 2 >1:39) = 0·0808 + [1-0.9172] = 0·0828 +0.08 28 =0·1657 70 = 0.3889 H₂: M3.5 =.0.31 n=25 X = 3.6 3.6-3.5 0.4 Pos 5. Every year, the Hometown Hardware Store expects an average of $606.40 per local family to be spent on springtime home repairs. A new analyst in the store disputes this and challenges that the average spending per local family will differ from this amount. She conducts a test based on a random sample of 30 households, using a 90% confidence level, resulting in a sample mean of $622.85. This is normally distributed using the population standard deviation of $65. Zear 600.85-806-40 Ho: μ = $606.40 Ha: #606.40 α=0.10 Z crit= 1.98 = 1.25 Z stat= -0.31 x= 0.05 S = 0.4 b = 3.5 2 crit= 1.645 P-vauve= P(21.05) = 0.1117 Zstat = 1.39 Decision: Decision: H. failed to be rejected 65 30 a = 0.10 a = One tail pvalue = 0.3783 6. A university wants to promote the graduates of its honors program. At present it is stated that the GPA of the graduates is 3.5 or below. The university wants to prove at a 90% confidence that the mean GPA of these graduates is greater than 3.5 A sample of 25 university honors program students is taken and is found to have a mean GPA equal to 3.6) The sample standard deviation is 0.40 5! Ha: M>3-5 2 crit= 2.952 tail 0.10 tail |pvalue = 0.1657 Reject the nusu hypothesis 2 a = 0.05 = 16.45 stat = 1.06 Decision: fail to reject Ho 11.86730 = 1.3860 1.39 7. Questions 4, 5, and 6 all have 90% confidence level. Why are the crit values different for all three? pvalue = 01117 Ĵ
Transcribed Image Text:7:36 PM Fri Nov 17 Tt O 4. It has been accepted that 40% or less of households changed their driving habits hat 40% because of escalating gas prices. However, a popular energy company challenges that % less than 40% of households have changed their driving habits because of escalating gas prices. 28. of 180 at 70 hous Their recent survey of 180 households finds that 70 households have made lifestyle changes due to escalating gas prices. (hint: first calculate the sample proportion) At a 10% level of significance, test the energy company's challenge. Ho: 0.40 Ha: p<0.40 α = 0.10 po= 40% n=180 X: 70 -1.28 0 p= X = 2080² 器 Z= 0.3889-0.40 0.3889x(1-03889) 180 P = P(Z < - 1:39) + ( 2 >1:39) = 0·0808 + [1-0.9172] = 0·0828 +0.08 28 =0·1657 70 = 0.3889 H₂: M3.5 =.0.31 n=25 X = 3.6 3.6-3.5 0.4 Pos 5. Every year, the Hometown Hardware Store expects an average of $606.40 per local family to be spent on springtime home repairs. A new analyst in the store disputes this and challenges that the average spending per local family will differ from this amount. She conducts a test based on a random sample of 30 households, using a 90% confidence level, resulting in a sample mean of $622.85. This is normally distributed using the population standard deviation of $65. Zear 600.85-806-40 Ho: μ = $606.40 Ha: #606.40 α=0.10 Z crit= 1.98 = 1.25 Z stat= -0.31 x= 0.05 S = 0.4 b = 3.5 2 crit= 1.645 P-vauve= P(21.05) = 0.1117 Zstat = 1.39 Decision: Decision: H. failed to be rejected 65 30 a = 0.10 a = One tail pvalue = 0.3783 6. A university wants to promote the graduates of its honors program. At present it is stated that the GPA of the graduates is 3.5 or below. The university wants to prove at a 90% confidence that the mean GPA of these graduates is greater than 3.5 A sample of 25 university honors program students is taken and is found to have a mean GPA equal to 3.6) The sample standard deviation is 0.40 5! Ha: M>3-5 2 crit= 2.952 tail 0.10 tail |pvalue = 0.1657 Reject the nusu hypothesis 2 a = 0.05 = 16.45 stat = 1.06 Decision: fail to reject Ho 11.86730 = 1.3860 1.39 7. Questions 4, 5, and 6 all have 90% confidence level. Why are the crit values different for all three? pvalue = 01117 Ĵ
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