Borrowing Problem 1’s story, the manufacturer of said light bulbs wants to deter

Question

Borrowing Problem 1’s story, the manufacturer of said light bulbs wants to determine an appropriate duration for a warranty. The decision depends on the validity of the following statement: our light bulb’s mean time to failure is greater than the industrystandard’s mean for similar light bulbs, which is 37.0833 weeks. Using the original information in Problem 1, conduct a hypothesis test as prompted below.

Resuming from Problem 3, a customer wants to buy many of these light bulbs. Unlike the manufacturer, the customer is more interested to know if the light bulb’s mean time to failure is at the industry-standard’s mean. This allows the customer to appropriately budget for the purchase (meaning, the mean shouldn’t be too low, which suggests poor quality bulb lights, nor should the mean be too high, which suggests more money than necessary would be spent). In addition, the customer does not know the standard deviation time to failure. However, the customer has information on the “36 light bulbs study”, including the sample mean and sample variance (from part 1e). Conduct a hypothesis test for this scenario. If any component(s) of the test is the same as in Problem 3, save some time and write “Same as #3” instead.

mean is 40

sample mean is 37.0833

standard deviation 11

sample variance is 121

Explanation / Answer

Solution:-

mean is 40

sample mean is 37.0833

standard deviation 11

sample variance is 121

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: = 40

Alternative hypothesis: 40

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 1.8333

DF = n - 1 = 36 - 1

D.F = 35

t = (x - ) / SE

t = - 1.591

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the t statistic having 35 degrees of freedom is less than -1.59 or greater than 1.59.

Thus, the P-value = 0.1206

Interpret results. Since the P-value (0.1206) is greater than the significance level (0.05), we cannot reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that light bulb’s mean time to failure is 40.

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