a manufacturer of bicycle tire has developed a new system which he claims an average life span of 5 years. with the standard deviation of 1.2 years a dealer product claims that the average life span of 150 samples is only 3.5 years. test the difference of population and sample means at 5% level of significance. what is the answer?

Do you really mean samples? If so, what is the n for each sample? If you really mean n, use information below.

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.

I need an answer

To test the difference between the population mean and sample mean at a 5% significance level, we can perform a one-sample t-test. The null hypothesis (H0) assumes that there is no difference between the population mean and the sample mean, while the alternative hypothesis (Ha) assumes that there is a significant difference.

Given:

- Population mean (μ) = 5 years
- Standard deviation (σ) = 1.2 years
- Sample mean (x̄) = 3.5 years
- Sample size (n) = 150
- Significance level (α) = 0.05

We can use the formula for the one-sample t-test statistic:

t = (x̄ - μ) / (σ / √n)

Calculating the t statistic:

t = (3.5 - 5) / (1.2 / √150)
t = -1.5 / (1.2 / √150)
t ≈ -1.5 / (1.2 / 12.25)
t ≈ -1.5 / 0.0984
t ≈ -15.24

To determine the critical t-value at a 5% significance level (two-tailed test) with (n-1) = 150-1 = 149 degrees of freedom, we can consult a t-distribution table or use statistical software. The critical t-value is approximately ± 1.976.

Since the calculated t-value (-15.24) is much larger than the critical t-value (±1.976), we can reject the null hypothesis (H0). This means there is a significant difference between the population mean and the sample mean.

Therefore, the dealer's claim of an average lifespan of 3.5 years is significantly different from the manufacturer's claim of an average lifespan of 5 years.

To test the difference between the population mean and the sample mean at a 5% level of significance, we can use a hypothesis test.

Step 1: State the Hypotheses
The null hypothesis (H0) states that there is no significant difference between the population mean and the sample mean. In this case, it would be H0: μ = 5 (population mean).

The alternative hypothesis (Ha) states that there is a significant difference between the population mean and the sample mean. In this case, it would be Ha: μ ≠ 5 (population mean).

Step 2: Set the Significance Level
The significance level (α) represents the probability of rejecting the null hypothesis when it is true. In this case, the significance level is 5% or 0.05.

Step 3: Compute the Test Statistic
We need to calculate the z-score (standardized test statistic) using the formula:

z = (x̄ - μ) / (σ / √n)

Where:
x̄ is the sample mean (3.5 years),
μ is the population mean (5 years),
σ is the standard deviation (1.2 years),
n is the sample size (150).

Substituting the values into the formula:

z = (3.5 - 5) / (1.2 / √150)

Step 4: Determine the Critical Value
Since α = 0.05 and it's a two-tailed test (not specified in the question, assuming it here), we need to divide the significance level by 2: 0.05 / 2 = 0.025.

Using a z-table or a statistical calculator, find the critical z-value for a significance level of 0.025 (considering it's a two-tailed test). Let's assume it as zα/2.

Step 5: Make a Decision
If the absolute value of the calculated z statistic is greater than the critical value (|z| > |zα/2|), we reject the null hypothesis. If it is not, we fail to reject the null hypothesis.

Step 6: Calculate the p-value
If needed, you can calculate the p-value associated with the test statistic. The p-value represents the probability of obtaining a test statistic as extreme as (or more extreme than) the observed data under the null hypothesis.

For a two-tailed test, we would find the probability of the test statistic being less than the negative z-value and greater than the positive z-value.

Step 7: Interpret the Results
Based on the decision made in Step 5, interpret the results in the context of the problem. If the null hypothesis is rejected, it means there is evidence to support the claim of a significant difference between the population mean and the sample mean.

Note: Since the critical value and p-value were not given, I cannot provide a concrete answer. You will need to perform the calculations and interpret the results based on the specific values.