Q1(a) A random sample of size n is drawn from normalpopulation with mean 5 and v

Question

Q1(a) A random sample of size n is drawn from normalpopulation with mean 5 and variance 100, If n=25, and t=2, what isvalues of ?
Q1(b) In a random sample of 200 persons having theirlunch at the University cafeteria on meatless day it was observedthat 30 percent preferred vegetable dishes. i. Find 95% confidence interval for p.
ii. How large a sample is needed if we want to be 95% confidentthat our estimate of is within 0.02?

Q1(b) In a random sample of 200 persons having theirlunch at the University cafeteria on meatless day it was observedthat 30 percent preferred vegetable dishes. i. Find 95% confidence interval for p.
ii. How large a sample is needed if we want to be 95% confidentthat our estimate of is within 0.02?

Explanation / Answer

Q. 1. (a) Is thereany Relationship between confidence interval and tests ofHypothesis?

      Justifyyour answer.

Answer:

There is an extremely closerelationship between confidence intervals and hypothesis testing.When a 95% confidence interval is constructed, all values in theinterval are considered plausible values for the parameter beingestimated. Values outside the interval are rejected as relativelyimplausible. If the value of the parameter specified by the nullhypothesis is contained in the 95% interval then the nullhypothesis cannot be rejected at the 0.05 level. If the valuespecified by the null hypothesis is not in the interval then thenull hypothesis can be rejected at the 0.05 level. If a 99%confidence interval is constructed, then values outside theinterval are rejected at the 0.01 level.

Q. 1. (b) A randomsample of size n is drawn from normal population with mean 5 andvariance 100, If n=25, and t=2, what is values of x-bar ?

Answer:

We know that, this is the questionof t-distribution, and is given below:

t = (x-bar –o) / (s / n)

Here, we have

o = 5

t = 2

s2 = 100

n = 25

s = s2 =100 = 10

t = (x-bar –o) / (s / n)

2 = (x-bar – 5) / (10 /25)

2 = (x-bar – 5) / (10 /5)

2 = (x-bar – 5) / (2)

2 * 2 = (x-bar – 5)

X-bar = 4 + 5

X-bar = 9

Q. 1. (c) In arandom sample of 200 persons having their lunch at the Universitycafeteria on meatless day it was observed that 30 percent preferredvegetable dishes.

i.         Find 95%confidence interval for p.

ii.       How large a sample isneeded if we want to be 95% confident that our estimate ofp is within 0.02?

Answer:

Here, we have;

Sample Size (n) = 200

Proportion of people whopreferred vegetable dishes (p) = 0.3

Proportion of people who didnot like vegetable dishes (q) = 1 – p = 1 – 0.3 =0.7

(i) Find 95% confidenceinterval for p

At 95% confidence, z from thez table = 1.96

Hence confidence interval (C.I.) = p ± z (pq /n)

Confidence Interval (C.I.) for p = 0.3 ± 1.96 (0.3 * 0.7 / 200)

Confidence Interval (C.I.) for p = 0.3 ± 1.96 (0.21 / 200)

Confidence Interval (C.I.) for p = 0.3 ± 1.96 0.00105

Confidence Interval (C.I.) for p = 0.3 ± 1.96 *0.032404

Confidence Interval (C.I.) for p = 0.3 ± 0.06351184

Confidence Interval (C.I.) for p = 0.236488, 0.363512 orfrom 0.2365 to 0.3635

(ii) How large a sample is needed if we want to be 95%confident that our estimate of p iswithin 0.02?

We know that;

Standard Deviation (S.D.) = (pq / n)

0.02 = (0.3 * 0.7 / n)

0.02 = (0.21 / n)

By taking squares both sides, we get

[0.02] 2 = [ (0.21 / n)] 2

0.0004 = (0.21 / n)

n = (0.21 / 0.0004)

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