The Quality Control Manager At A Light Bulb Factory Needs to

The Quality Control Manager At A Light Bulb Factory Needs to Determine The quality control manager at a light bulb factory needs to determine whether the mean life of a large shipment of light bulbs is equal to 375 hours. The population standard deviation is 100 hours. A random sample of 64 light bulbs indicates a sample mean life of 350 hours. a. What are the null and alternate hypotheses? b. Draw a diagram labeling all axes and the rejection & failing to reject regions. c. At the .05 level of significance, what is your conclusion? Why? d. Suppose a Type II error occurs. What might be the consequences of the event and who suffers? e. Consider a population with a known standard deviation of 26.8. In order to compute an interval estimate for the population mean, a sample of 64 observations is drawn.

- Is the condition that the population is normally distributed satisfied? Explain.

- Compute the margin of error at a 95% confidence level.

- Compute the margin of error at a 95% confidence level based on a larger sample of 225 observations.

- Which of the two margins of error will lead to a wider confidence interval? The margin of error with 6.3% confidence or the margin of error with 99% confidence?

Paper For Above instruction In this analysis, we examine the hypothesis testing process concerning the mean life of light bulbs and the implications of statistical errors. The fourth paragraph presents the specific scenario: a light bulb manufacturer tests whether the average lifespan of their shipment aligns with 375 hours, a manufacturing quality benchmark. The sample data comprise 64 light bulbs with a mean lifespan of 350 hours, and the population standard deviation is known to be 100 hours. a. The null hypothesis (H0) posits that the population mean lifespan µ is equal to 375 hours (H0: µ = 375).

The alternative hypothesis (H1) suggests that µ is not equal to 375 hours (H1: µ ≠ 375). This is a two-tailed test aimed at determining whether the sample provides enough evidence to reject the stated population mean. b. To visualize the hypothesis test, one would construct a normal distribution curve centered at the

hypothesized mean of 375 hours. The rejection regions would be located in the tails where the z-scores exceed the critical values for a significance level of α = 0.05. For a two-tailed test, the critical z-values are approximately ±1.96. Regions beyond these z-scores represent where the null hypothesis would be rejected, indicating the sample mean is significantly different from 375 hours. c. The test statistic (z) can be calculated using the formula: z = (X■ - µ) / (σ / √n), where X■ is the sample mean, µ is the hypothesized mean, σ is the population standard deviation, and n is the sample size.

Plugging in the values: z = (350 - 375) / (100 / √64) = (-25) / (100 / 8) = (-25) / 12.5 = -2.0.

Since |z| = 2.0 is greater than the critical value of 1.96, we reject the null hypothesis at the 0.05 significance level. The conclusion is that there is statistically significant evidence to suggest the mean lifespan differs from 375 hours. d. A Type II error occurs if the null hypothesis is not rejected when it is false. In this context, it means failing to detect that the true mean lifespan is not 375 hours when it actually isn't. The consequence of this error could be that the factory continues production based on the incorrect assumption that lamps meet quality standards, potentially leading to customer dissatisfaction, increased warranty claims, and damage to the manufacturer's reputation. The consumers (end-users) and the company both suffer from such a mistake, with consumers receiving substandard products and the manufacturer facing financial and reputational losses. e. For the second part, with a population standard deviation of 26.8 and a sample size of 64, the condition that the population is normally distributed is generally satisfied if either the population distribution is normal or the sample size is large enough (n ≥ 30) by the Central Limit Theorem to approximate normality. Since n=64 exceeds 30, the sample mean distribution can be considered approximately normal, making the use of z-based confidence intervals appropriate.

The margin of error (ME) at a 95% confidence level is calculated as: ME = z* (σ / √n), where z* for 95% confidence is approximately 1.96.

ME = 1.96 * (26.8 / √64) = 1.96 * (26.8 / 8) = 1.96 * 3.35 ≈ 6.57.

For a larger sample of 225 observations, the margin of error becomes:

ME = 1.96 * (26.8 / √225) = 1.96 * (26.8 / 15) = 1.96 * 1.79 ≈ 3.51.

The margin of error with the larger sample (225 observations) is narrower (about 3.51) compared to the smaller sample's margin (about 6.57). Consequently, the larger sample provides a more precise estimate of the population mean.

Regarding the confidence levels, the confidence interval with 99% confidence uses a larger z* value of approximately 2.576, leading to a wider margin of error: ME = 2.576 * (26.8 / √64) = 2.576 * 3.35 ≈ 8.63.

Since the margin of error at 99% confidence (8.63) is larger than at 95% confidence (6.57), the interval with 99% confidence will be wider. The 6.3% confidence statement appears to be a mislabel or typo; the general consideration is that higher confidence levels produce wider intervals, ensuring higher certainty but less precision.

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