Instructions Clearly label each problem and each part of a problem on your paper
ID: 3066618 • Letter: I
Question
Instructions Clearly label each problem and each part of a problem on your paper. Make sure and print information and explain in detail on any questions or comments. The following data set represents the repair costs (in dollars) for a random sample of 24 dishwashers. {Note: the data listed below should be entered into one column.) 41.82 52.8178.16 83.4878.88 1.) 88.13 88.79 90.07 90.35 91.72 95.43 96.50 101.32 105.59 105.62 .32 7.14 118.42 118.77 119.01 30.70 141.52 141.84 147.06 Use MINTAB to construct a 85% confidence interval for the population mean. {Stat/Basic Statistics/1-sample t! 2.) In MINITAB generate 100 columns of data with 100 observations in each column from a Normal distribution with -46ando-5 { Calc Random DatalNormal . Find an appropriate 92% confidence interval for the population mean using each of the 100 samples. {Stat/Basic Statistics/1-sample zj a.)What are the 100 confidence intervals (MINTAB output)? b.)How many of these intervals are expected to contain the mean of the population? c.)Based on your Minitab output, how many of these intervals actually contain the mean of the population)?Explanation / Answer
Que.1
MTB > OneT C1;
SUBC> Confidence 85;
SUBC> Alternative 0.
One-Sample T: C1
Variable N Mean StDev SE Mean 85% CI
C1 24 101.48 25.93 5.29 (93.59, 109.36)
Comment:
85% confidence interval for population mean is (93.59, 109.36)
Que.3
MTB > OneT C1;
SUBC> Test 225;
SUBC> Confidence 99;
SUBC> Alternative 1.
One-Sample T: C1
Test of ? = 225 vs > 225
Variable N Mean StDev SE Mean 99% Lower Bound T P
C1 20 249.2 52.8 11.8 219.2 2.05 0.027
Comment:
Since p-value is greater than 0.01, hence we accept null hypothesis at 1% l.o.s.
That is mean value of the purchases of cardholders in a month is not increased its same as 225.
Que.4
a)
Regression Analysis: y versus x
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
Regression 1 10240 10240.4 10.58 0.031
x 1 10240 10240.4 10.58 0.031
Error 4 3872 967.9
Total 5 14112
Model Summary
S R-sq R-sq(adj) R-sq(pred)
31.1112 72.56% 65.71% 12.77%
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant 227.8 26.0 8.76 0.001
x -2.856 0.878 -3.25 0.031 1.00
Regression Equation
y = 227.8 - 2.856 x
b)
Correlation: y, x
Pearson correlation of y and x = -0.852
Comment: There is high degree of negative correlation between y and x. It means that if value of y increases then value of x decreases with greater extent.
c)
Coefficient of determination:
R2= 72.56%
It means that 72.56 % of variation in y is explained by x.
d) Regression equation:
y = 227.8 - 2.856 x
e)
We cannot predict for x = 145. Because this value is far from the range of x which is used to fit the regression line.
f) If x=30
y= 227.8 - 2.856*30 = 142.12