Question

Recording times (in minutes) for a sample of Dr. Marion's 33 recordings of Gustav Mahler's 1st Symphony:

55 49 50 52 53 56 55 57 59 52 55 50 50 62 53 58

58 63 51 52 51 54 65 60 57 57 56 54 56 47 48 52

1. Estimate the mean recording time for Mahler’s 1st Symphony.

2. To draw any inferences from these data, what must we assume and why?

3. Find a 95% confidence interval for the mean recording time of all recordings of Mahler’s 1st Symphony.

4. Find a 90% cinfudence interval for Mahler’s 1st Symphony mean recording time. Which of your two intervalis wider, and why?

5. What hypothesis would you use to test that the mean recording time for Mahler’s 1st Symphony is significantly longer than 52 minutes? Define in words any parameters you use in your hypotheses.

6. What test-statistic should you use to test these hyptheses? What is the sampling distribution of this test-statistic if Null H0 is true?

7. Find the P-Value for this test-statistic, state your conclusion, interpret the result, and compare your result to the interval in “a.” for the approximate value α.

8. If your friendly local statistics professor informs you that the σ ≈ 4 for all recordings of Mahler’s 1st Symphony, how does this change your answers in b-g?

• I. What must we assume about our sampling population, and why?

• II. Find a 95% confidence interval for the mean recording time.

• III. Find and compare a 90% confidence interval for the mean recording time.

• IV. What hypothesis would you use to test that the mean recording time for Mahler’s 1st Symphony is significantly longer than 52 minutes

• V. What test-statistic should you use to test these hyptheses?

• VI. What is the sampling distribution of this test-statistic under Null H0?

• VII. Find the P-Value, state, and interpret your conclusions at ∝ = 0.02 .

• VIII. Explain how and why you test results change.

9. Using σ = 4 and your sample mean, estimate the probabilities that:

• I. Mean recording time of the 64 recordings exceeds 55 minutes.

• II. Total recording time of 36 recording exceeds 30 hours.

10. Using σ = 4, how large a sample would we need in order to estimate mean recording time for Mahler’s 1st Symphony to a:

• I. 95% error margin of 3 minutes?

• II. 90% error margin of 2 minutes?

• III. Compare your answers and explain their difference.

Answer 1

We are given the following recording times of Gustav Mahler's 1st Symphony:

55,49,50,52,53,56,55,57,59,52,,55,50,50,62,53,58,58,63,51,52,51,54,65,60,57,57,56,54,56,47,48,52.

We have to estimate the mean recording time for Mahler's 1st Symphony.

Mean recording time of Symphony = 49.82(sum of all observations/no.of observations)

To draw any inferences from the data, we need to make the following assumptions:

1)Since sample size is large(>30), the sample comes from a population which follows normal distribution with parameters and . We make this assumption because when sample size is greater than 30 it becomes a large sample and follows normal distribution. From the central limit theorem, we know that if our sample is large, the sampling distriution will be approximately normally distributed irrespective of the shape of the population distribution.

2)The samples are randomly sampled from a well defined population.

We now have to find 95% confidence interval of mean recording time of all recordings of Mahler's first Symphony.

Let be a random sample drawn from normal population with mean and variance .

x~N(0,1)

We know that ~

Consider standard normal distribution for

From standard normal tables at 5% level of significance,

Therefore 95% confidence interval for the population mean for the normal distribution are

We now find 90% confidence interval

Let be a random sample drawn from normal population with mean and variance .

x~N(0,1)

We know that ~

Consider standard normal distribution for

From standard normal tables at 10% level of significance,

=0.90

Therefore 90% confidence interval for confidence interval for the population mean for the normal distribution are

The 95% confidence interval is wider because it has a lower critical value for lower level of confidence which decreases the margin of error. Therefore 95% confidence interval is wider than the 90% confidence level.