the manufacturer of a certain brand of auto batteries claims the mean life of these batteries is 45 months. A consumer protection agency that wants to check this claim took a random sample of 36 such batteries and found the mean life for this sample is 43.75 months with a standard deviation of 4.5 months. Find the p-value for the test hypothesis with the alternative hypothesis that the mean life of these batteries is less than 45 months.
Would you reject the manufactures claim at a 5% significance level? Explain
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 that Z score. Is it less than .05?
standard deviation is 4months
Z=(43.75-45)/(4.5/sqrt36)=-1.67
p-value=0.0475
the manufacture claim is mu=45 months
p-value=2*0.0475=0.095
alpha=0.05
the p-value > alpha
therefore we fail to reject the null alternative
To answer this question, we need to perform a hypothesis test.
Let's define the null hypothesis (H0) and alternative hypothesis (Ha):
H0: The mean life of these batteries is 45 months.
Ha: The mean life of these batteries is less than 45 months.
We have a sample of 36 batteries with a sample mean of 43.75 months and a standard deviation of 4.5 months. To test the hypothesis, we can use the one-sample t-test because we know the sample mean and standard deviation, but not the population parameters.
First, let's calculate the t-statistic:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
t = (43.75 - 45) / (4.5 / sqrt(36))
t = (-1.25) / (4.5 / 6)
t = -1.25 / 0.75
t ≈ -1.67
Next, we need to find the p-value associated with this t-statistic. The p-value represents the probability of observing a t-statistic as extreme as or more extreme than the observed value, assuming the null hypothesis is true.
Since we have a one-sided alternative hypothesis that the mean life is less than 45 months, we need to find the area under the t-distribution curve to the left of the observed t-statistic.
We can consult a t-distribution table or use statistical software to find the p-value. Assuming a significance level of 5%, the p-value can be compared to this threshold.
In this case, the p-value is approximately 0.056. Since the p-value is greater than the significance level of 0.05, we fail to reject the null hypothesis.
Therefore, based on the given data, we do not have sufficient evidence to reject the manufacturer's claim that the mean life of these batteries is 45 months at a 5% significance level.