Question

Hello experts. does the square of the standard deviation or average(mean) make a change in the calculation? Please show your work

Labels on 3.79 litre cans of paint usually indicate the drying time and the area that can be covered in one coat. Most brands indicate that, in one coat, 3.79 L will cover between 23.2 and 46.4 square metre, depending on the texture of the surface to be painted. One manufacturer claims that 3.79L of its paint will cover 37.2 square metre of surface area. A sample of ten 3.79L cans of white paint were used to paint ten identical areas using the same kind of equipment, The average was 33.39 square metre and the standard deviation 4.49 square metre.

A. Find the 90% confidence interval for the mean area covered by this company’s paint.

B. At α =0.10 is there enough evidence to indicate the average coverage is different from the company’s claim? Use both the rejection point and p-value methods

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Answer 1

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 by this 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 to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 37.2
Alternative hypothesis: u 37.2

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 1.4199

DF = n - 1

D.F = 9
t = (x - u) / SE

t = - 2.68

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

Critical method

D.F = 9

t_critical = 1.833

Rejection region is - 1.833 > t > 1.833 Interpret results. 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 probability that the t statistic having 9 degrees of freedom is less than -2.68 or greater than 2.68.

Thus, the P-value = 0.025

Interpret results. Since the P-value (0.025) is less than the significance level (0.10), we have to reject the null hypothesis.