Use for #1-2. A researcher wished to estimate the difference between the proport

Question

Use for #1-2. A researcher wished to estimate the difference between the proportions of users of two shampoos who are satisfied with the product. In a sample of 400 users of Shampoo A taken by this researcher, 78 said they are satisfied. In another sample of 500 users of Shampoo B taken by the same researcher, 92 said they were satisfied.

1. Construct a 90% confidence interval for the true difference between the two population proportions.

2. Test the claim that there is no difference in customer satisfaction between the two shampoos. Use a significance level of 0.01.

Use for #3-5. A survey recently reported that the mean annual expenditure for inpatient and outpatient services of a random sample of 365 persons over age 64 living in Hudson was $5,423 with a standard deviation of $979. A random sample of 352 persons over age 64 living in Sudbury had an average expense of $5,516 with a standard deviation of $882.

3. Can we conclude that the mean inpatient and outpatient expense of all Sudbury residents over age 64 is higher than for the residents of Hudson? Report a p-value and use a 0.01 level of significance.

4. Would your answer have changed if you had used a .05 level of significance? Explain.

5. Find a 99% confidence interval for the difference in expenditure for inpatient and outpatient services.

Explanation / Answer

Q1.
Confidence Interval for Diffrence of Proportion
CI = (p1 - p2) ± Z a/2 Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 )
Proportion 1
No. of chances( X1 )=78
No.Of Observed (n1)=400
P1= X1/n1=0.195
Proportion 2
No. of chances(X2)=92
No.Of Observed (n2)=500
P2= X2/n2=0.184
C.I = (0.195-0.184) ±Z a/2 * Sqrt( (0.195*0.805/400) + (0.184*0.816/500) )
=(0.195-0.184) ± 1.64* Sqrt(0.001)
=0.011-0.043,0.011+0.043
=[-0.032,0.054]

Q2.
Given that,
sample one, x1 =78, n1 =400, p1= x1/n1=0.195
sample two, x2 =92, n2 =500, p2= x2/n2=0.184
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.01
from standard normal table, two tailed z /2 =
since our test is two-tailed
reject Ho, if zo < -2.576 OR if zo > 2.576
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.195-0.184)/sqrt((0.189*0.811(1/400+1/500))
zo =0.419
| zo | =0.419
critical value
the value of |z | at los 0.01% is 2.576
we got |zo| =0.419 & | z | =2.576
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.4189 ) = 0.6753
hence value of p0.01 < 0.6753,here we do not reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 0.419
critical value: -2.576 , 2.576
decision: do not reject Ho
p-value: 0.6753

we have proof that no difference in customer satisfaction between the two shampoos

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