Consumers own laptops of brand X for a certain amount of time before buying a new model. This time duration has a distribution with unknown mean and standard deviation 1.5. The company X expects that consumers will keep their laptop or 3.5 years. After collecting a sample of 27 customers they calculate a sample mean of 4 years. Is there enough evidence (95% significance) to support the claim that laptops are kept for more than 3.5 years?
I will ask my friend she knows statistics! also she should reply soon.
Nevermind she was busy sorry!
Please only post your questions once. Repeating posts will not get a quicker response. In addition, it wastes our time looking over reposts that have already been answered in another post. Thank you.
See your later post.
Jooby, if you can't provide an answer to the problem, do not respond.
To determine whether there is enough evidence to support the claim that laptops are kept for more than 3.5 years, we can perform a hypothesis test.
First, let's state the null and alternative hypotheses:
Null Hypothesis (H0): The mean time duration before buying a new laptop is equal to or less than 3.5 years.
Alternative Hypothesis (Ha): The mean time duration before buying a new laptop is more than 3.5 years.
Next, we need to calculate the test statistic. In this case, we can use a one-sample t-test since the population mean and standard deviation are unknown. The formula for the t-test statistic is:
t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))
Given the information provided, the sample mean (x̄) is 4 years, the hypothesized mean (μ) is 3.5 years, the sample standard deviation (s) is 1.5 years, and the sample size (n) is 27.
Using these values, we can calculate the test statistic as follows:
t = (4 - 3.5) / (1.5 / sqrt(27))
Simplifying the equation:
t = 0.5 / (1.5 / 5.2)
t = 0.5 / 0.577
t ≈ 0.867
Now, we need to determine the critical value for the t-distribution with a 95% significance level and 26 degrees of freedom (sample size - 1) using a t-table or statistical software. For a one-tailed test, the critical value is approximately 1.706.
Finally, we compare the calculated test statistic (0.867) with the critical value (1.706) to assess whether there is enough evidence to support the alternative hypothesis. Since the calculated test statistic does not exceed the critical value, we fail to reject the null hypothesis.
Therefore, based on the given sample data, there is not enough evidence (at the 95% significance level) to support the claim that laptops are kept for more than 3.5 years.