Statistics

[jsmith241182]

I have 3 questions that i need to answer and i really cant get my head round them for some reason and i would appreciate it very much if anyone could help me. The questions are:

In order to estimate the percentage of males who smoke in bed to within 2% for a 95% confidence a pilot survey was carried out. Of a random sample of 35 males, 5 were found to smoke in bed. How many males need to be sampled in the main survey ?


A sample of grain is inspected for its purity. The mean percentage of impurities 5%, the standard deviation is 2%. Assuming that the percentage of impurities is normally distributed, determine the proportion of specimens which have a impurity level of (a). less than 5%, (b). below 10% and (c) above 10%.


Determine if a correlation exists between the following data: District Proportion of Open Space Proportion of Accidents (%) to children as a % of all accidents A 5.0 46.3 B 2.2 43.4 C 1.3 42.9 D 4.2 42.2 E 1.4 40.0 F 2.0 38.8 G 7.0 38.2 H 2.5 37.4 I 4.5 37.0 J 3.1 33.3 K 5.2 33.6 L 7.2 33.6 M 6.3 30.8 N 12.2 28.3 O 14.6 23.8 P 23.6 17.8 Q 14.8 17.1 R 27.5 10.8

I would be so grateful for the help. Cheers

[Rassis]

Question 1:

n = 35; p = 5/35; e = 0.02
n0 = [Z(a/2)]^2.[p(1 – p)]/e^2 = 1.95996^2 x 5/35 x (1 – 5/35)/0.02^2 = 1,176

Question 2:

mean = 5%; stdev = 2%

Using the expression Z = (p - mean) / stdev, and a table of mutual conversion of Z and P, you get:

P(p <= 5%) = 0.5
P(p <= 10%) = 0.99379
P(p > 10%) = 1 – 0.99379 = 0.00621

Question 3:

Yes, there is sufficient evidence of correlation (inverse) between the two variables.

Pearson coefficient = -0.912
R^2 = 0.8315
Spearman coefficient = -0.811

I leave the proof to you.

[Rassis]

I corrected the result of the first question above: it must be 1,176 instead of 2,401 as I wrote the first place. I am so sorry.

[jsmith241182]

Thankyou very much for your help. Its much appreciated