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

MATLAB: An Introduction with Applications
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Only Question 7 i need the answer 

 
 
 
### Problem 4
It has been accepted that 40% or less of households changed their driving habits because of escalating gas prices. However, a popular energy company challenges that more than 40% of households have changed their driving habits because of escalating gas prices.

Their recent survey of 180 households finds that 70 households have made lifestyle changes due to escalating gas prices. At a 10% level of significance, test the energy company's challenge.

#### Hypotheses
- \( H_0: p \leq 0.40 \)
- \( H_a: p > 0.40 \)

#### Calculations
- \( x = 70 \)
- \( n = 180 \)
- \( \hat{p} = \frac{70}{180} = 0.3889 \)
- \( z = \frac{0.3889 - 0.40}{\sqrt{\frac{0.40 \times 0.60}{180}}} = -0.31 \)

#### Normal Curve Representation
The curve shows a normal distribution with the rejection region marked in the right tail. The critical z-value (\( Z_{crit} \)) is 1.28 for a significance level (\( \alpha \)) of 0.10. 

#### Decision
\( Z_{stat} = -0.31 \). Since \( Z_{stat} < Z_{crit} \), fail to reject \( H_0 \).
- p-value = 0.3783

### Problem 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.

A test is conducted on a random sample of 30 households, using a 90% confidence level, resulting in a sample mean of $632.85. The spending is normally distributed with a population standard deviation of $65.

#### Hypotheses
- \( H_0: \mu = 606.40 \)
- \( H_a: \mu \neq 606.40 \)

#### Calculations
- \( \mu = 632.85 \)
- \( \sigma = 65 \)
- \( n = 30 \)
- Standard Error = \( \frac{65}{\sqrt{30}} \)
- \( z
Transcribed Image Text:### Problem 4 It has been accepted that 40% or less of households changed their driving habits because of escalating gas prices. However, a popular energy company challenges that more than 40% of households have changed their driving habits because of escalating gas prices. Their recent survey of 180 households finds that 70 households have made lifestyle changes due to escalating gas prices. At a 10% level of significance, test the energy company's challenge. #### Hypotheses - \( H_0: p \leq 0.40 \) - \( H_a: p > 0.40 \) #### Calculations - \( x = 70 \) - \( n = 180 \) - \( \hat{p} = \frac{70}{180} = 0.3889 \) - \( z = \frac{0.3889 - 0.40}{\sqrt{\frac{0.40 \times 0.60}{180}}} = -0.31 \) #### Normal Curve Representation The curve shows a normal distribution with the rejection region marked in the right tail. The critical z-value (\( Z_{crit} \)) is 1.28 for a significance level (\( \alpha \)) of 0.10. #### Decision \( Z_{stat} = -0.31 \). Since \( Z_{stat} < Z_{crit} \), fail to reject \( H_0 \). - p-value = 0.3783 ### Problem 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. A test is conducted on a random sample of 30 households, using a 90% confidence level, resulting in a sample mean of $632.85. The spending is normally distributed with a population standard deviation of $65. #### Hypotheses - \( H_0: \mu = 606.40 \) - \( H_a: \mu \neq 606.40 \) #### Calculations - \( \mu = 632.85 \) - \( \sigma = 65 \) - \( n = 30 \) - Standard Error = \( \frac{65}{\sqrt{30}} \) - \( z
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