Question

3. A simple random sample of n = 37 bolts was taken from a shipment resulting in a sample

mean diameter of x= 1.82 cm and a sample standard deviation s = 0.26 cm. Use an a = 0.05
significance level to test the claim that the mean diameter is equal to 1.75 cm.

Answer 1

To test the claim that the mean diameter is equal to 1.75 cm, a one-sample t-test can be conducted. The test results show that there is not enough evidence to support the claim that the mean diameter is different from 1.75 cm.

Step-by-step explanation:

To test the claim that the mean diameter is equal to 1.75 cm, we can conduct a one-sample t-test using the given sample mean and standard deviation.

Step 1: State the null and alternative hypotheses:

Null hypothesis (H0): The mean diameter is equal to 1.75 cm.

Alternative hypothesis (Ha): The mean diameter is not equal to 1.75 cm.

Step 2: Set the significance level:

Given: α = 0.05

Step 3: Calculate the test statistic:

Using the formula for the t-statistic: t = (x - μ) / (s / √n)

Where x = sample mean, μ = hypothesized population mean, s = sample standard deviation, and n = sample size.

Plugging in the values: t = (1.82 - 1.75) / (0.26 / √37) ≈ 1.04

Step 4: Determine the critical value(s):

Since this is a two-tailed test, we need to find the critical values that correspond to the significance level of α/2 (0.05/2 = 0.025) in the t-distribution with (n-1) degrees of freedom (37-1 = 36). Checking the t-table or using statistical software, we find that the critical values are approximately ±2.0301.

Step 5: Make a decision:

If the absolute value of the test statistic is greater than the critical value(s), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

In this case, the absolute value of the test statistic (1.04) is less than the critical value (2.0301). Therefore, we fail to reject the null hypothesis.

Conclusion: Based on the sample evidence, there is not enough evidence to support the claim that the mean diameter is different from 1.75 cm at a significance level of 0.05.

Answer 2

There is insufficient evidence at the 5% level of significance to reject H₀​. This suggests that, based on the sample, the mean diameter is not significantly different from 1.75 cm. This implies that the claim of a mean diameter of 1.75 cm is supported by the available data.

Step-by-step explanation:

Bolts have an mean diameter (μ) of 1.75 cm, with a standard deviation of 0.26 cm. Assuming the diameter, X, is normally distributed:

`X \sim \text{N}(\mu, \sigma^2)=X\sim \text{N}(1.75, 0.26^2)`

In a sample of 37 bolts, the mean diameter (
`\overline{x}`) is 1.82 cm.

To test the claim that the mean diameter is equal to 1.75 cm, we can perform a hypothesis test.

The null hypothesis (H₀​) is that the mean diameter is equal to 1.75 cm, and the alternative hypothesis (H₁​) is that the mean diameter is not equal to 1.75 cm. Therefore, the hypotheses are as follows:

`H_0: \;\;\mu = 1.75`

`H_1:\;\; \mu \\eq 1.75`

The sample mean
`\overline{x}` is 1.82 where n = 37.

Therefore, the test statistic is:

`z=\frac{\overline{x}-\mu}{\sigma / √(n)}=(1.82-1.75)/(0.26/√(37)) \approx 1.63766683...`

This is a two-tailed test.

Given α = 0.05, the critical region is given by:

`\text{P}(Z > z) = (\alpha)/(2) = 0.025,\;\;\sf or`

`\text{P}(Z < -z) = (\alpha)/(2) = 0.025`

Using a calculator, this gives a value for z of 1.95996402...

So the critical region is:

`Z > 1.9599... \;\;\textsf{or}\;\; Z < -1.9599...`

Since z = 1.6376... the observed value of the test statistic does not lie in the critical region and therefore is not significant.

There is insufficient evidence at the 5% level of significance to reject H₀ in favour of the alternative hypothesis that the mean diameter is not equal to 1.75 cm. In other words, we do not have enough evidence to claim that the mean diameter is different from 1.75 cm based on the given sample.