An extensive survey in 1970 revealed preschool children spend an average of 6.3 hours per day watching television. The distribution of TV times in normal with a population standard deviation of 2 hours. Last year a sample of 100 preschool children had a mean of 5.8 hours of TV watching per day. Test to determine if the average time preschool children spend watching TV per day has decreased at the 0.01 significance level. Part A- Hoi u =
An extensive survey in 1970 revealed preschool children spend an average of 6.3 hours per day watching television. The distribution of TV times in normal with a population standard deviation of 2 hours. Last year a sample of 100 preschool children had a mean of 5.8 hours of TV watching per day. Test to determine if the average time preschool children spend watching TV per day has decreased at the 0.01 significance level. Part A- Hoi u =
MATLAB: An Introduction with Applications
6th Edition
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Author:Amos Gilat
Publisher:Amos Gilat
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Need D through G
![### Question 8
An extensive survey in 1970 revealed preschool children spend an average of 6.3 hours per day watching television. The distribution of TV times is normal with a population standard deviation of 2 hours. Last year a sample of 100 preschool children had a mean of 5.8 hours of TV watching per day. Test to determine if the average time preschool children spend watching TV per day has decreased at the 0.01 significance level.
**Part A - \(H_0: \mu =\)**
\[ \mu = 6.3 \text{ hours} \]
**Part B - \(H_a: \mu\)**
\[ \mu < 6.3 \text{ hours} \]
**Part C - State alpha**
\[ \alpha = 0.01 \]
**Part D - Reject null if**
\[ z < -z_\alpha \]
**Part E - The test statistic is**
\[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]
\[ z = \frac{5.8 - 6.3}{2 / \sqrt{100}} \]
\[ z = \frac{-0.5}{0.2} \]
\[ z = -2.5 \]
**Part F - Write the decision**
\[ \text{Since } z = -2.5 < -z_{0.01} = -2.33, \text{ we reject the null hypothesis.} \]
**Part G - At \(\alpha = 0.01\) , there is evidence to**
Reject the null hypothesis. This suggests that the average time preschool children spend watching TV per day has significantly decreased since 1970.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff5fb381f-811c-41ea-92d9-ccdf3c6fabee%2F3ae2a2c2-6f2a-4ee8-be96-0e00e94ce28b%2F4ytx3t8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 8
An extensive survey in 1970 revealed preschool children spend an average of 6.3 hours per day watching television. The distribution of TV times is normal with a population standard deviation of 2 hours. Last year a sample of 100 preschool children had a mean of 5.8 hours of TV watching per day. Test to determine if the average time preschool children spend watching TV per day has decreased at the 0.01 significance level.
**Part A - \(H_0: \mu =\)**
\[ \mu = 6.3 \text{ hours} \]
**Part B - \(H_a: \mu\)**
\[ \mu < 6.3 \text{ hours} \]
**Part C - State alpha**
\[ \alpha = 0.01 \]
**Part D - Reject null if**
\[ z < -z_\alpha \]
**Part E - The test statistic is**
\[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]
\[ z = \frac{5.8 - 6.3}{2 / \sqrt{100}} \]
\[ z = \frac{-0.5}{0.2} \]
\[ z = -2.5 \]
**Part F - Write the decision**
\[ \text{Since } z = -2.5 < -z_{0.01} = -2.33, \text{ we reject the null hypothesis.} \]
**Part G - At \(\alpha = 0.01\) , there is evidence to**
Reject the null hypothesis. This suggests that the average time preschool children spend watching TV per day has significantly decreased since 1970.
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