Question 3 (a) What is the difference between confidence interval and hypothesis

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Question 3 (a) What is the difference between confidence interval and hypothesis testing? (b) To estimate the mean amount spent per were collected for a sample of 49 customers Assunne a population standard deviation of customer for dinner at r restaurant, daia m S5, (i) At 95% confidence, what is the margin of error? (ii) If the sample mean is S24So, what is the 95% confidence interval for the population mean? (c) A manufacturer claimed that at least 95% of the machine equipment he supplied to factory conformed to the specifications. An examination of a sample of 200 pieces of equipment revealed that 18 were faulty. Test his claim at a significance level of 0.01 Question 4 A semiconductor industry would like to find out whether worker hours are related to lot sizes. They wish to use the model to later predict the number of worker hours required for different lot sizes. The observed values are given below Lot 20 20 30 304040 50 50 60 60 70 70 80 80 size, 0 | 55 | 73 | 67 | 87 | 95 | 108 | 112 | 128 | 135 | 148 | 160 | 170 | 162 Given x-700, 1550, 2 = 40600, y-193402, xy- 88480. (a) Find the estimated regression line to fit the data using the method of least squares. (b) Interpret , the slope of the regression line. (c) Predict the number of worker hours needed for a lot of size 75.

Explanation / Answer

Q3.
a.
both methods are used to estimate the sample data behaviour when compared to population. Hypothesis is well structured and works with the predefined mean values to concludes the decision to accept or reject.
confidence level is approximate range of values that we assume to mean is lies in the interval. This is useful to know the magnitude of an effect on sample

b.
given that,
i.
standard deviation, =5
sample mean, x =58.3
population size (n)=49
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 5/ sqrt ( 49) )
= 0.71
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 0.71
= 1.4

ii.
given that,
standard deviation, =5
sample mean, x =24.8
population size (n)=49
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 24.8 ± Z a/2 ( 5/ Sqrt ( 49) ) ]
= [ 24.8 - 1.96 * (0.71) , 24.8 + 1.96 * (0.71) ]
= [ 23.4,26.2 ]
-----------------

c.
Given that,
possibile chances (x)=182
sample size(n)=200
success rate ( p )= x/n = 0.91
success probability,( po )=0.95
failure probability,( qo) = 0.05
null, Ho:p>=0.95  
alternate, H1: p<0.95
level of significance, = 0.01
from standard normal table,left tailed z /2 =2.33
since our test is left-tailed
reject Ho, if zo < -2.33
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.91-0.95/(sqrt(0.0475)/200)
zo =-2.5955
| zo | =2.5955
critical value
the value of |z | at los 0.01% is 2.33
we got |zo| =2.596 & | z | =2.33
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: left tail - Ha : ( p < -2.59554 ) = 0.00472
hence value of p0.01 > 0.00472,here we reject Ho
ANSWERS
---------------
null, Ho:p>=0.95
alternate, H1: p<0.95
test statistic: -2.5955
critical value: -2.33
decision: reject Ho
p-value: 0.00472

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