Solved on Jan 05, 2024

Valve regulates water pressure on 180 engines. Mean pressure is 6.2 lbs/square inch, variance is 0.25. Determine if there is sufficient evidence at 0.02 level that valve does not meet 6.3 lbs/square inch specification.
$H_0$ : $\mu = 6.3$ lbs/square inch $H_a$ : $\mu \neq 6.3$ lbs/square inch

STEP 1

Assumptions
1. The sample size (number of engines tested) is $n = 180$ .
2. The sample mean pressure is $\bar{x} = 6.2$ lbs/square inch.
3. The population variance is known to be $\sigma^2 = 0.25$ .
4. The population mean pressure that the valve is designed to produce is $\mu_0 = 6.3$ lbs/square inch.
5. The significance level for the test is $\alpha = 0.02$ .
6. We are conducting a two-tailed test to determine if there is evidence that the valve does not perform to specifications.

STEP 2

State the null and alternative hypotheses.
The null hypothesis ( $H_0$ ) is that the mean pressure produced by the valve is equal to the designed mean pressure. The alternative hypothesis ( $H_a$ ) is that the mean pressure produced by the valve is not equal to the designed mean pressure.
$H_0: \mu = \mu_0$ $H_a: \mu \neq \mu_0$

STEP 3

Calculate the standard error of the mean. The standard error of the mean (SEM) is the standard deviation of the sampling distribution of the sample mean. It is calculated by dividing the population standard deviation by the square root of the sample size.
$SEM = \frac{\sigma}{\sqrt{n}}$

STEP 4

Plug in the values for the population standard deviation ( $\sigma = \sqrt{0.25} = 0.5$ ) and the sample size ( $n = 180$ ) to calculate the SEM.
$SEM = \frac{0.5}{\sqrt{180}}$

STEP 5

Calculate the SEM.
$SEM = \frac{0.5}{\sqrt{180}} \approx \frac{0.5}{13.4164} \approx 0.0373$

STEP 6

Calculate the test statistic (z-score). The test statistic is calculated by subtracting the population mean under the null hypothesis from the sample mean, and then dividing the result by the SEM.
$z = \frac{\bar{x} - \mu_0}{SEM}$

STEP 7

Plug in the values for the sample mean ( $\bar{x} = 6.2$ ), the population mean under the null hypothesis ( $\mu_0 = 6.3$ ), and the SEM ( $\approx 0.0373$ ) to calculate the z-score.
$z = \frac{6.2 - 6.3}{0.0373}$

STEP 8

Calculate the z-score.
$z = \frac{6.2 - 6.3}{0.0373} \approx \frac{-0.1}{0.0373} \approx -2.68$

STEP 9

Determine the critical z-value(s) for a two-tailed test at the $\alpha = 0.02$ significance level. Since it is a two-tailed test, we will split the alpha level into two, so each tail will have an area of $\alpha/2 = 0.01$ .

STEP 10

Using a standard normal distribution table or a z-score calculator, find the critical z-value that corresponds to the upper and lower 1% of the distribution.
For $\alpha/2 = 0.01$ , the critical z-values are approximately $\pm2.33$ .

STEP 11

Compare the calculated z-score to the critical z-values. If the calculated z-score falls outside the range defined by the critical z-values, we reject the null hypothesis.

STEP 12

Since our calculated z-score of approximately $-2.68$ is less than the lower critical z-value of $-2.33$ , we have sufficient evidence at the $0.02$ level to reject the null hypothesis.

STEP 13

Conclude the hypothesis test.
There is sufficient evidence at the 0.02 significance level to conclude that the valve does not perform to the specifications of producing a mean pressure of $6.3$ lbs/square inch.

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Valve regulates water pressure on 180 engines. Mean pressure is 6.2 lbs/square inch, variance is 0.25. Determine if there is sufficient evidence at 0.02 level that valve does not meet 6.3 lbs/square inch specification.
$H_0$ : $\mu = 6.3$ lbs/square inch $H_a$ : $\mu \neq 6.3$ lbs/square inch

STEP 1

STEP 2

STEP 3

Calculate the standard error of the mean. The standard error of the mean (SEM) is the standard deviation of the sampling distribution of the sample mean. It is calculated by dividing the population standard deviation by the square root of the sample size.
$SEM = \frac{\sigma}{\sqrt{n}}$

STEP 4

Plug in the values for the population standard deviation ( $\sigma = \sqrt{0.25} = 0.5$ ) and the sample size ( $n = 180$ ) to calculate the SEM.
$SEM = \frac{0.5}{\sqrt{180}}$

STEP 5

Calculate the SEM.
$SEM = \frac{0.5}{\sqrt{180}} \approx \frac{0.5}{13.4164} \approx 0.0373$

STEP 6

Calculate the test statistic (z-score). The test statistic is calculated by subtracting the population mean under the null hypothesis from the sample mean, and then dividing the result by the SEM.
$z = \frac{\bar{x} - \mu_0}{SEM}$

STEP 7

Plug in the values for the sample mean ( $\bar{x} = 6.2$ ), the population mean under the null hypothesis ( $\mu_0 = 6.3$ ), and the SEM ( $\approx 0.0373$ ) to calculate the z-score.
$z = \frac{6.2 - 6.3}{0.0373}$

STEP 8

Calculate the z-score.
$z = \frac{6.2 - 6.3}{0.0373} \approx \frac{-0.1}{0.0373} \approx -2.68$

STEP 9

Determine the critical z-value(s) for a two-tailed test at the $\alpha = 0.02$ significance level. Since it is a two-tailed test, we will split the alpha level into two, so each tail will have an area of $\alpha/2 = 0.01$ .

STEP 10

Using a standard normal distribution table or a z-score calculator, find the critical z-value that corresponds to the upper and lower 1% of the distribution.
For $\alpha/2 = 0.01$ , the critical z-values are approximately $\pm2.33$ .

STEP 11

Compare the calculated z-score to the critical z-values. If the calculated z-score falls outside the range defined by the critical z-values, we reject the null hypothesis.

STEP 12

Since our calculated z-score of approximately $-2.68$ is less than the lower critical z-value of $-2.33$ , we have sufficient evidence at the $0.02$ level to reject the null hypothesis.

STEP 13

Conclude the hypothesis test.
There is sufficient evidence at the 0.02 significance level to conclude that the valve does not perform to the specifications of producing a mean pressure of $6.3$ lbs/square inch.

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STEP 1

STEP 2

STEP 3

STEP 4

Plug in the values for the population standard deviation (σ=0.25=0.5\sigma = \sqrt{0.25} = 0.5σ=0.25​=0.5) and the sample size (n=180n = 180n=180) to calculate the SEM.SEM=0.5180SEM = \frac{0.5}{\sqrt{180}}SEM=180​0.5​

STEP 5

Calculate the SEM.SEM=0.5180≈0.513.4164≈0.0373SEM = \frac{0.5}{\sqrt{180}} \approx \frac{0.5}{13.4164} \approx 0.0373SEM=180​0.5​≈13.41640.5​≈0.0373

STEP 6

Calculate the test statistic (z-score). The test statistic is calculated by subtracting the population mean under the null hypothesis from the sample mean, and then dividing the result by the SEM.z=xˉ−μ0SEMz = \frac{\bar{x} - \mu_0}{SEM}z=SEMxˉ−μ0​​

STEP 7

Plug in the values for the sample mean (xˉ=6.2\bar{x} = 6.2xˉ=6.2), the population mean under the null hypothesis (μ0=6.3\mu_0 = 6.3μ0​=6.3), and the SEM (≈0.0373\approx 0.0373≈0.0373) to calculate the z-score.z=6.2−6.30.0373z = \frac{6.2 - 6.3}{0.0373}z=0.03736.2−6.3​

STEP 8

Calculate the z-score.z=6.2−6.30.0373≈−0.10.0373≈−2.68z = \frac{6.2 - 6.3}{0.0373} \approx \frac{-0.1}{0.0373} \approx -2.68z=0.03736.2−6.3​≈0.0373−0.1​≈−2.68

STEP 9

Determine the critical z-value(s) for a two-tailed test at the α=0.02\alpha = 0.02α=0.02 significance level. Since it is a two-tailed test, we will split the alpha level into two, so each tail will have an area of α/2=0.01\alpha/2 = 0.01α/2=0.01.

STEP 10

Using a standard normal distribution table or a z-score calculator, find the critical z-value that corresponds to the upper and lower 1% of the distribution.For α/2=0.01\alpha/2 = 0.01α/2=0.01, the critical z-values are approximately ±2.33\pm2.33±2.33.

STEP 11

Compare the calculated z-score to the critical z-values. If the calculated z-score falls outside the range defined by the critical z-values, we reject the null hypothesis.

STEP 12

Since our calculated z-score of approximately −2.68-2.68−2.68 is less than the lower critical z-value of −2.33-2.33−2.33, we have sufficient evidence at the 0.020.020.02 level to reject the null hypothesis.

STEP 13

Conclude the hypothesis test.There is sufficient evidence at the 0.02 significance level to conclude that the valve does not perform to the specifications of producing a mean pressure of 6.36.36.3 lbs/square inch.

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Plug in the values for the population standard deviation ( $\sigma = \sqrt{0.25} = 0.5$ ) and the sample size ( $n = 180$ ) to calculate the SEM.
$SEM = \frac{0.5}{\sqrt{180}}$

Calculate the SEM.
$SEM = \frac{0.5}{\sqrt{180}} \approx \frac{0.5}{13.4164} \approx 0.0373$

Calculate the test statistic (z-score). The test statistic is calculated by subtracting the population mean under the null hypothesis from the sample mean, and then dividing the result by the SEM.
$z = \frac{\bar{x} - \mu_0}{SEM}$

Plug in the values for the sample mean ( $\bar{x} = 6.2$ ), the population mean under the null hypothesis ( $\mu_0 = 6.3$ ), and the SEM ( $\approx 0.0373$ ) to calculate the z-score.
$z = \frac{6.2 - 6.3}{0.0373}$

Calculate the z-score.
$z = \frac{6.2 - 6.3}{0.0373} \approx \frac{-0.1}{0.0373} \approx -2.68$

Determine the critical z-value(s) for a two-tailed test at the $\alpha = 0.02$ significance level. Since it is a two-tailed test, we will split the alpha level into two, so each tail will have an area of $\alpha/2 = 0.01$ .

Using a standard normal distribution table or a z-score calculator, find the critical z-value that corresponds to the upper and lower 1% of the distribution.
For $\alpha/2 = 0.01$ , the critical z-values are approximately $\pm2.33$ .

Since our calculated z-score of approximately $-2.68$ is less than the lower critical z-value of $-2.33$ , we have sufficient evidence at the $0.02$ level to reject the null hypothesis.

Conclude the hypothesis test.
There is sufficient evidence at the 0.02 significance level to conclude that the valve does not perform to the specifications of producing a mean pressure of $6.3$ lbs/square inch.