Hello experts. does the square
Hello experts. does the square of the standard deviation oraverage(mean) make a change in the calculation? Please show yourwork
Labels on 3.79 litre cans of paint usually indicate the dryingtime and the area that can be covered in one coat. Most brandsindicate that, in one coat, 3.79 L will cover between 23.2 and 46.4square metre, depending on the texture of the surface to bepainted. One manufacturer claims that 3.79L of its paint will cover37.2 square metre of surface area. A sample of ten 3.79L cans ofwhite paint were used to paint ten identical areas using the samekind of equipment, The average was 33.39 square metre and thestandard deviation 4.49 square metre.
A. Find the 90% confidence interval for the mean area covered bythis company’s paint.
B. At α =0.10 is there enough evidence to indicate the averagecoverage is different from the company’s claim? Use both therejection point and p-value methods
.
Answer:
Solution:-
The square of the standard deviation or average(mean)make a change in the calculation.
A) 90% confidence interval for the mean area covered bythis company’s paint is C.I = ( 30.787, 35.993).
C.I = 33.39 +1.833 × 1.4199
C.I = 33.39 +2.603
C.I = ( 30.787, 35.993)
B)
State the hypotheses. The first step is tostate the null hypothesis and an alternative hypothesis.
Null hypothesis: u = 37.2Alternative hypothesis: u 
37.2
Note that these hypotheses constitute a two-tailed test. Thenull hypothesis will be rejected if the sample mean is too big orif it is too small.
Formulate an analysis plan. For this analysis,the significance level is 0.10. The test method is a one-samplet-test.
Analyze sample data. Using sample data, wecompute the standard error (SE), degrees of freedom (DF), and the tstatistic test statistic (t).
SE = s / sqrt(n)
S.E = 1.4199
DF = n – 1
D.F = 9t = (x – u) / SE
t = – 2.68
where s is the standard deviation of the sample, x is the samplemean, u is the hypothesized population mean, and n is the samplesize.
Critical method
D.F = 9
tcritical = 1.833
Rejection region is – 1.833 > t > 1.833 Interpretresults. Since the t-value (-2.68) lies in the rejection region,hence we have to reject the null hypothesis.
P-value method
Since we have a two-tailed test, the P-value is the probabilitythat the t statistic having 9 degrees of freedom is less than -2.68or greater than 2.68.
Thus, the P-value = 0.025
Interpret results. Since the P-value (0.025) is lessthan the significance level (0.10), we have to reject the nullhypothesis.
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