Question

1. A study was done to measure the length of a full grown adult St. Bernard dog. It is assumed that the lengths of these types of dogs follow a N(μ, σ ) distribution where the length is measured in inches. For a random sample of 25 dogs, a 95% confidence interval for μ was calculated to be (42.16, 57.84).

   1. (a) Provide a clear interpretation of what this interval telling us. (1 mark)

   2. (b) The researcher feels that this interval is too wide and wants to reduce it to a width at most 12 units.

      Assuming that the value of σ is unknown, answer the following questions.

      1. (i) For a confidence level of 95%, calculate the smallest sample size needed.

      2. (ii) For a sample size fixed at n = 25, what confidence level 100(1 − α)% is needed? For this question you may not be able to get an exact answer, so just state between what two confidence levels would your confidence interval be.

   3. (c) For this question, assume that based on historical data, the value of the population standard deviation is known to be σ = 20. Using the same sample size of 25 and assuming the 95% confidence interval is still (42.16, 57.84), what would be am 83% confidence interval for μ?

Answer 1

Sample size = n = 25

95% Confidence interval for population mean length of dog () is given by -

95% CI = [42.16, 57.84]

a) Hence, we are 95% confident that the mean length of full grown adult St. Bernard dog will lie between 42.16 inches and 57.84 inches.

b) 95% CI = [42.16, 57.84]

Hence,

= [42.16, 57.84]

As the critical value of t with 24 degrees of freedom for two tailed test at 0.05 level of significance is 2.064.

= [ 42.16, 57.84]

So,

_____eqn (1)

______eqn(2)

Adding eqn (1) and (2), we get,

_________ eqn (3)

And, 2.064 * s/5 = 57.84 - 50

0.4128 * s = 7.84

s = 18.992

Now, Reseacher wants to reduce atmost 12 units.

UPPER LIMIT OF 95% CI - LOWER LIMIT OF 95% CI = 12

ii) Since, n is fixed at 25, let at level of significance we get width of CI atmost 12 units. Then,

Using t table at 24 degrees of freedom we will look for probability for which t value is close to 1.58 and it will be between 0.05 and 0.10 for one tailed test and between 0.10 and 0.20 for two tailed test.

Hence, confidence level for this will be between 80% and 90%.

c) Standard deviation = = 20

Sample size = n = 25

95% CI = [42.16, 57.84]

Level of Significance = 1 - 0.83 = 0.17

Hence, sample mean = = 50 (as calculated above in eqn (3))

Now, since, the standard deviation is Known, mean will follow the z distribution

So, 83% CI for is given by -

Now the critical value of z for two tailed test at 0.83 level of significance can be obtained using function=NORMSINV(0.17/2) in excel and we will take absolute value of that.

It will be equal to 1.37.

So, z critical= 1.37

Hence, 83% CI will be given by -

= [44.52, 55.48]