Thank you very much! The topic is about statistical hypothesis. 93. Bricks. A pu

Question

Thank you very much! The topic is about statistical hypothesis.

93. Bricks. A purchaser of bricks suspects that the quality of bricks is deterio- rating. From past experience, the mean crushing strength of such bricks is 400 pounds. A sample of n = 100 bricks yielded a mean of 395 pounds and standard deviation of 20 pounds. (a) Test the hypothesis that the mean quality has not changed against the alternative that it has deteriorated. Choose .05. (b) What is the p-value for the test in (a) (e) Assume that the producer of the bricks contested your findings in (a) and (b). Their company suggested constructing the 95% confidence interval for with a total length of no more than 4. what sample size is needed to construct such a confidence interval? 94 Soybeans. According to advertisements, a strain of soybeans planted on oil prepared with a specific fertilizer treatment has a mean yield of 500 bushels per acre. Fifty farmers who belong to a cooperative plant the soy- beans. Each uses a 40-acre plot and records the mean yield per acre. The mean and variance for the sample of 50 farms are 485 and s10045. Use the p-value for this test to determine whether the data provide suffi- cient evidence to indicate that the mean yield for the soybeans is different from that advertised

Explanation / Answer

Solution:-

9.4)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: = 500
Alternative hypothesis: 500

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 0.05. 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 = 14.174

DF = n - 1 = 50 - 1

D.F = 49
t = (x - ) / SE

t = - 1.06

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

Since we have a two-tailed test, the P-value is the probability that the t statistic having 49 degrees of freedom is less than -1.06 or greater than 1.06.

Thus, the P-value = 0.2943

Interpret results. Since the P-value (0.2943) is greater than the significance level (0.05), we cannot reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that the mean yield for the soyabeans is different from that advertised.

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