Question A manufacturer claims the mean breaking strength of new cables is greater than 1775 lbs. Test this claim using a one-tailed test with $\alpha=0.05$ . The sample mean is 1790 lbs and the population standard deviation is 55 lbs.
(a) $H_0: \mu \leq 1775$ lbs, $H_1: \mu > 1775$ lbs (b) Use a z-test (c) Test statistic $z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} = 1.818$ (d) $p$ -value $= 0.035$ (e) Yes, we can support the claim that the mean breaking strength is greater than 1775 lbs.

Studdy Solution

STEP 1

Assumptions
1. The historical mean breaking strength of the cables is 1775 pounds.
2. The historical standard deviation of the breaking strength is 55 pounds.
3. The sample size of newly manufactured cables is 18.
4. The sample mean breaking strength of the newly manufactured cables is 1790 pounds.
5. The population is assumed to be normally distributed.
6. The population standard deviation is assumed to have not changed and is still 55 pounds.
7. The level of significance for the test is 0.05.
8. We are performing a one-tailed test to determine if the mean breaking strength has increased.

STEP 2

State the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$ .
$H_{0}: \mu = 1775 \text{ pounds} \\ H_{1}: \mu > 1775 \text{ pounds}$

STEP 3

Determine the type of test statistic to use.
Since we are given the population standard deviation and the sample size is relatively small, we will use the z-test for the test statistic.

STEP 4

Find the value of the test statistic.
The test statistic for a z-test is calculated using the formula:
$z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$
where $\bar{x}$ is the sample mean, $\mu$ is the population mean under the null hypothesis, $\sigma$ is the population standard deviation, and $n$ is the sample size.

STEP 5

Plug in the values to calculate the test statistic.
$z = \frac{1790 - 1775}{\frac{55}{\sqrt{18}}}$

STEP 6

Calculate the denominator of the test statistic.
$\frac{55}{\sqrt{18}} = \frac{55}{4.243} \approx 12.962$

STEP 7

Calculate the test statistic.
$z = \frac{1790 - 1775}{12.962} \approx \frac{15}{12.962} \approx 1.157$

STEP 8

Round the test statistic to three decimal places.
$z \approx 1.157$

STEP 9

Find the $p$ -value.
The $p$ -value is the probability of observing a test statistic as extreme as, or more extreme than, the one observed if the null hypothesis is true. For a one-tailed test, we look up the $p$ -value corresponding to our z-value in the standard normal distribution (z-table).

STEP 10

Using a z-table or a statistical software, find the $p$ -value that corresponds to the z-value of 1.157.
The $p$ -value for z = 1.157 is approximately 0.1235. However, since we are conducting a one-tailed test, we need the area to the right of the z-value.

STEP 11

Calculate the one-tailed $p$ -value.
$p\text{-value} = 1 - 0.1235 \approx 0.8765$

STEP 12

Round the $p$ -value to three decimal places.
$p\text{-value} \approx 0.876$

STEP 13

Make a decision regarding the null hypothesis.
Since the $p$ -value (0.876) is greater than the level of significance (0.05), we do not reject the null hypothesis.

STEP 14

State the conclusion.
We do not have sufficient evidence at the 0.05 level of significance to support the claim that the population mean breaking strength of the newly-manufactured cables is greater than 1775 pounds.

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Studdy Solution

STEP 1

STEP 2

State the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$ .
$H_{0}: \mu = 1775 \text{ pounds} \\ H_{1}: \mu > 1775 \text{ pounds}$

STEP 3

Determine the type of test statistic to use.
Since we are given the population standard deviation and the sample size is relatively small, we will use the z-test for the test statistic.

STEP 4

STEP 5

Plug in the values to calculate the test statistic.
$z = \frac{1790 - 1775}{\frac{55}{\sqrt{18}}}$

STEP 6

Calculate the denominator of the test statistic.
$\frac{55}{\sqrt{18}} = \frac{55}{4.243} \approx 12.962$

STEP 7

Calculate the test statistic.
$z = \frac{1790 - 1775}{12.962} \approx \frac{15}{12.962} \approx 1.157$

STEP 8

Round the test statistic to three decimal places.
$z \approx 1.157$

STEP 9

Find the $p$ -value.
The $p$ -value is the probability of observing a test statistic as extreme as, or more extreme than, the one observed if the null hypothesis is true. For a one-tailed test, we look up the $p$ -value corresponding to our z-value in the standard normal distribution (z-table).

STEP 10

Using a z-table or a statistical software, find the $p$ -value that corresponds to the z-value of 1.157.
The $p$ -value for z = 1.157 is approximately 0.1235. However, since we are conducting a one-tailed test, we need the area to the right of the z-value.

STEP 11

Calculate the one-tailed $p$ -value.
$p\text{-value} = 1 - 0.1235 \approx 0.8765$

STEP 12

Round the $p$ -value to three decimal places.
$p\text{-value} \approx 0.876$

STEP 13

Make a decision regarding the null hypothesis.
Since the $p$ -value (0.876) is greater than the level of significance (0.05), we do not reject the null hypothesis.

STEP 14

State the conclusion.
We do not have sufficient evidence at the 0.05 level of significance to support the claim that the population mean breaking strength of the newly-manufactured cables is greater than 1775 pounds.

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Studdy Solution

STEP 1

STEP 2

State the null hypothesis H0H_{0}H0​ and the alternative hypothesis H1H_{1}H1​.H0:μ=1775 poundsH1:μ>1775 pounds H_{0}: \mu = 1775 \text{ pounds} \\ H_{1}: \mu > 1775 \text{ pounds} H0​:μ=1775 poundsH1​:μ>1775 pounds

STEP 3

Determine the type of test statistic to use.Since we are given the population standard deviation and the sample size is relatively small, we will use the z-test for the test statistic.

STEP 4

STEP 5

Plug in the values to calculate the test statistic.z=1790−17755518 z = \frac{1790 - 1775}{\frac{55}{\sqrt{18}}} z=18​55​1790−1775​

STEP 6

Calculate the denominator of the test statistic.5518=554.243≈12.962 \frac{55}{\sqrt{18}} = \frac{55}{4.243} \approx 12.962 18​55​=4.24355​≈12.962

STEP 7

Calculate the test statistic.z=1790−177512.962≈1512.962≈1.157 z = \frac{1790 - 1775}{12.962} \approx \frac{15}{12.962} \approx 1.157 z=12.9621790−1775​≈12.96215​≈1.157

STEP 8

Round the test statistic to three decimal places.z≈1.157 z \approx 1.157 z≈1.157

STEP 9

Find the ppp-value.The ppp-value is the probability of observing a test statistic as extreme as, or more extreme than, the one observed if the null hypothesis is true. For a one-tailed test, we look up the ppp-value corresponding to our z-value in the standard normal distribution (z-table).

STEP 10

Using a z-table or a statistical software, find the ppp-value that corresponds to the z-value of 1.157.The ppp-value for z = 1.157 is approximately 0.1235. However, since we are conducting a one-tailed test, we need the area to the right of the z-value.

STEP 11

Calculate the one-tailed ppp-value.p-value=1−0.1235≈0.8765 p\text{-value} = 1 - 0.1235 \approx 0.8765 p-value=1−0.1235≈0.8765

STEP 12

Round the ppp-value to three decimal places.p-value≈0.876 p\text{-value} \approx 0.876 p-value≈0.876

STEP 13

Make a decision regarding the null hypothesis.Since the ppp-value (0.876) is greater than the level of significance (0.05), we do not reject the null hypothesis.

STEP 14

State the conclusion.We do not have sufficient evidence at the 0.05 level of significance to support the claim that the population mean breaking strength of the newly-manufactured cables is greater than 1775 pounds.

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State the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$ .
$H_{0}: \mu = 1775 \text{ pounds} \\ H_{1}: \mu > 1775 \text{ pounds}$

Determine the type of test statistic to use.
Since we are given the population standard deviation and the sample size is relatively small, we will use the z-test for the test statistic.

Plug in the values to calculate the test statistic.
$z = \frac{1790 - 1775}{\frac{55}{\sqrt{18}}}$

Calculate the denominator of the test statistic.
$\frac{55}{\sqrt{18}} = \frac{55}{4.243} \approx 12.962$

Calculate the test statistic.
$z = \frac{1790 - 1775}{12.962} \approx \frac{15}{12.962} \approx 1.157$

Round the test statistic to three decimal places.
$z \approx 1.157$

Find the $p$ -value.
The $p$ -value is the probability of observing a test statistic as extreme as, or more extreme than, the one observed if the null hypothesis is true. For a one-tailed test, we look up the $p$ -value corresponding to our z-value in the standard normal distribution (z-table).

Using a z-table or a statistical software, find the $p$ -value that corresponds to the z-value of 1.157.
The $p$ -value for z = 1.157 is approximately 0.1235. However, since we are conducting a one-tailed test, we need the area to the right of the z-value.

Calculate the one-tailed $p$ -value.
$p\text{-value} = 1 - 0.1235 \approx 0.8765$

Round the $p$ -value to three decimal places.
$p\text{-value} \approx 0.876$

Make a decision regarding the null hypothesis.
Since the $p$ -value (0.876) is greater than the level of significance (0.05), we do not reject the null hypothesis.

State the conclusion.
We do not have sufficient evidence at the 0.05 level of significance to support the claim that the population mean breaking strength of the newly-manufactured cables is greater than 1775 pounds.