Math / Data & Statistics

QuestionSuppose the weights of seventh-graders at a certain school vary according to a Normal distribution, with a mean of 100 poun and a standard deviation of 7.5 pounds. A researcher believes the average weight has decreased since the implementation of a new breakfast and lunch program at the school. She finds, in a random sample of 35 students, an average weight of 98 pounds.
What is the $P$ -value for an appropriate hypothesis test of the researcher's claim? $-1.578$ 0.115 0.943 0.057

Studdy Solution

STEP 1

1. The weights of seventh-graders are normally distributed.
2. The population mean weight before the program is $\mu = 100$ pounds.
3. The population standard deviation is $\sigma = 7.5$ pounds.
4. The sample size is $n = 35$ .
5. The sample mean weight is $\bar{x} = 98$ pounds.
6. We are conducting a one-sample z-test for the mean.

STEP 2

1. State the null and alternative hypotheses.
2. Calculate the test statistic (z-score).
3. Determine the P-value from the z-score.
4. Compare the P-value to the significance level.

STEP 3

State the null hypothesis ( $H_0$ ) and the alternative hypothesis ( $H_a$ ):
- $H_0: \mu = 100$ (The average weight has not decreased.) - $H_a: \mu < 100$ (The average weight has decreased.)

STEP 4

Calculate the test statistic (z-score) using the formula:
$z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$
Substitute the given values:
$z = \frac{98 - 100}{\frac{7.5}{\sqrt{35}}}$

STEP 5

Calculate the standard error:
$\text{Standard Error} = \frac{7.5}{\sqrt{35}} \approx 1.267$
Calculate the z-score:
$z = \frac{-2}{1.267} \approx -1.578$

STEP 6

Determine the P-value for $z = -1.578$ using a standard normal distribution table or calculator. The P-value corresponds to the probability that $Z$ is less than $-1.578$ .

STEP 7

The P-value for $z = -1.578$ is approximately $0.057$ .

STEP 8

Compare the P-value to the significance level (commonly $\alpha = 0.05$ ):
- If $P$ -value $< \alpha$ , reject $H_0$ . - If $P$ -value $\geq \alpha$ , do not reject $H_0$ .
Since $0.057 > 0.05$ , we do not reject $H_0$ .
The $P$ -value is:
$\boxed{0.057}$

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Studdy Solution

STEP 1

STEP 2

1. State the null and alternative hypotheses.
2. Calculate the test statistic (z-score).
3. Determine the P-value from the z-score.
4. Compare the P-value to the significance level.

STEP 3

State the null hypothesis ( $H_0$ ) and the alternative hypothesis ( $H_a$ ):
- $H_0: \mu = 100$ (The average weight has not decreased.) - $H_a: \mu < 100$ (The average weight has decreased.)

STEP 4

Calculate the test statistic (z-score) using the formula:
$z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$
Substitute the given values:
$z = \frac{98 - 100}{\frac{7.5}{\sqrt{35}}}$

STEP 5

Calculate the standard error:
$\text{Standard Error} = \frac{7.5}{\sqrt{35}} \approx 1.267$
Calculate the z-score:
$z = \frac{-2}{1.267} \approx -1.578$

STEP 6

Determine the P-value for $z = -1.578$ using a standard normal distribution table or calculator. The P-value corresponds to the probability that $Z$ is less than $-1.578$ .

STEP 7

The P-value for $z = -1.578$ is approximately $0.057$ .

STEP 8

Compare the P-value to the significance level (commonly $\alpha = 0.05$ ):
- If $P$ -value $< \alpha$ , reject $H_0$ . - If $P$ -value $\geq \alpha$ , do not reject $H_0$ .
Since $0.057 > 0.05$ , we do not reject $H_0$ .
The $P$ -value is:
$\boxed{0.057}$

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Studdy Solution

STEP 1

STEP 2

1. State the null and alternative hypotheses.2. Calculate the test statistic (z-score).3. Determine the P-value from the z-score.4. Compare the P-value to the significance level.

STEP 3

State the null hypothesis (H0H_0H0​) and the alternative hypothesis (HaH_aHa​):- H0:μ=100H_0: \mu = 100H0​:μ=100 (The average weight has not decreased.) - Ha:μ<100H_a: \mu < 100Ha​:μ<100 (The average weight has decreased.)

STEP 4

Calculate the test statistic (z-score) using the formula:z=xˉ−μσn z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} z=n​σ​xˉ−μ​Substitute the given values:z=98−1007.535 z = \frac{98 - 100}{\frac{7.5}{\sqrt{35}}} z=35​7.5​98−100​

STEP 5

Calculate the standard error:Standard Error=7.535≈1.267 \text{Standard Error} = \frac{7.5}{\sqrt{35}} \approx 1.267 Standard Error=35​7.5​≈1.267Calculate the z-score:z=−21.267≈−1.578 z = \frac{-2}{1.267} \approx -1.578 z=1.267−2​≈−1.578

STEP 6

Determine the P-value for z=−1.578z = -1.578z=−1.578 using a standard normal distribution table or calculator. The P-value corresponds to the probability that ZZZ is less than −1.578-1.578−1.578.

STEP 7

The P-value for z=−1.578z = -1.578z=−1.578 is approximately 0.0570.0570.057.

STEP 8

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1. State the null and alternative hypotheses.
2. Calculate the test statistic (z-score).
3. Determine the P-value from the z-score.
4. Compare the P-value to the significance level.

State the null hypothesis ( $H_0$ ) and the alternative hypothesis ( $H_a$ ):
- $H_0: \mu = 100$ (The average weight has not decreased.) - $H_a: \mu < 100$ (The average weight has decreased.)

Calculate the test statistic (z-score) using the formula:
$z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$
Substitute the given values:
$z = \frac{98 - 100}{\frac{7.5}{\sqrt{35}}}$

Calculate the standard error:
$\text{Standard Error} = \frac{7.5}{\sqrt{35}} \approx 1.267$
Calculate the z-score:
$z = \frac{-2}{1.267} \approx -1.578$

Determine the P-value for $z = -1.578$ using a standard normal distribution table or calculator. The P-value corresponds to the probability that $Z$ is less than $-1.578$ .

The P-value for $z = -1.578$ is approximately $0.057$ .