Question

Showing your work and including the appropriate diagrams where they should be obvious to include, determine the following and highlight your final answer:

a) The value of zα/2 when determine the 96% confidence interval for p.

b) The decision rule (aka, the rejection region) for testing the following pair of hypotheses at the .05 level of significance when the population standard deviation is unknown and a random sample of size 28 is taken. H0: µ = 18 Ha: µ < 18

c) The value of tα/2 when determine the 99% confidence interval for µ when the sample standard deviation of the population is unknown and a sample size of 35 was taken from a normally distribution random variable

d) P(Y > 57) where Y has a normal distribution with µ = 64 and σ = 7.

e) The decision rule (aka, the rejection region) for testing the following pair of hypotheses at the .01 level of significance when a sample size of 45 is taken. H0: p = .45 Ha: p ≠ .45

f) P ( < 0.35) where is approximately normally distributed with p = .33 and n = 100.

g) Determine the p-value for testing: H0: µ = 43 Ha: µ ≠ 43 when a random sample of size 30 was taken from a normal population whose standard deviation is known and the value of the test statistic equals -1.38.

h) P ( > 22) where Y has a normal distribution with a mean of 20, a standard deviation of 8, and, when a sample size of 25 was taken.  

i) The 90% confidence interval for the population mean when a random sample of size 16 was taken from a very large population and its mean was calculated to be 22 and its standard deviation was calculated to be 3. (The population had a normal distribution with an unknown standard deviation.) No concluding statement is required.

j) Determine the p-value for testing: H0: µ = 15 Ha: µ > 15 when a random sample of size 18 was taken from a normal population whose standard deviation is unknown and the value of the test statistic equals 2.35.

k) The 98% confidence interval for the population proportion of successes when a random sample of size 80 was taken from a very large population and the number of successes in that sample was counted to be 20. No concluding statement is required.  

Answer 1

a) Given, confidence level=96%

alpha = 0.04 , alpha /2 =0.04/2 = 0.02

Zalpha/2 = Z0.02= 2.053

b) Reject the Null Hypothesis if the value of test statistic is greater than Z table value otherwise our decision is failed to reject null hypothesis.

Using p value, reject the Null Hypothesis if p value is less than alpha otherwise accept the Null Hypothesis.

c) confidence level=99%

Alpha=0.01 , using Excel,

To find the alue of tα/2= TINV(1-Apha/2, degrees of freedom)= 2.0019

d) To find P(Y > 57) where Y has a normal distribution with µ = 64 and σ = 7.

Using Excel formula,

P(Y>57) = 1-NORMDIST(X=57, MEAN=64, SD=7, CUMULATIVE=TRUE)

P(Y>57) = 0.8413

e) similar to b).

f) To find P ( Y< 0.35) where is approximately normally distributed with p = .33 and n = 100.

Using binomial approximation to normal,

Mean == 0.33 and

Standard deviatideviation. = =

sigma = 0.047

P(Y<0.35) = NORMDIST(0.35,0.33,0.047,TRUE)

P(Y<0.35) = 0.6647

g)for testing: H0: µ = 43 Ha: µ ≠ 43 , the value of the test statistic equals -1.38. n=30.

Test statistic

Sigma   =(43*√30)/1.38 = 170.66

P value =1- NORMSINV (test statistic)= 0.08379

h) to find P ( Y> 22) where Y has a normal distribution with a mean of 20, a standard deviation of 8, and, when a sample size of 25 was taken. Using Excel,

P(Y>22)=1- NORMDIST (22,20,8,1)= 0.4012

i) Given, n= 16,

Z table=Z alpha = 1.28

The 90% confidence interval for the population mean is ,

= (22-1.28*3 , 22+1.28*3)

= (18.16 , 25.84 )

j) Test statistic= 2.35 , using Excel,

P value= 1- NORMSDIST (z)

Pvalue=1- NORMSDIST(2.35)= 0.0093

k) Given ,

n=80 , X=20 ,

Sample proportion= p=X/n = 20/80 = 0.25

Z table at alpha= 0.02=2.053.

98% confidence interval for population proportion is ,

= ( 0.25 - 2.053*0.0484 , 0.25 + 2.053 * 0.0484 )

= ( 0.150 , 0.349 )