Suppose we wish to estimate the average daily yield of a chemical manufactured i

Question

Suppose we wish to estimate the average daily yield of a chemical manufactured in a chemical plant. The daily yield, recorded for n=50 days, produced a mean and a standard deviation of tons and S = 21 tons. Use .

(a) Explain concretely what does mean alpha=0.05?. 5 marks

(b) What is the meaning of (1-alpha)= 0.95? 5 marks

(c) Test the hypothesis that the average daily yield of the chemical is tons per day against the alternative that is greater or less than 880 tons per day using the same sample information.

(d) Build the confidence interval for the mean.

Explanation / Answer

a)
Alpha = 0.05 indicates the significance level.
It is the probability that we reject the null hypothesis when it is true.

b)
0.95 represents the probability that we accept the null hypothesis.

In order to solve other two parts, mean is required. Please provide the data.

We need sample mean to solve other two parts.

c)

test statistics,

z = (871 - 880)/(21/sqrt(50))

z = -3.0305

p-value = 0.0024

as this is two tailed test we find the p-value as 2*P(z < -3.0305)

d)

CI for 95%

n = 50

mean = 871

z-value of 95% CI = 1.9600

std. dev. = 21

SE = std.dev./sqrt(n) = 2.96985

ME = z*SE = 5.8208

Lower Limit = Mean - ME = 865.1792

Upper Limit = Mean + ME = 876.8208

95% CI (865.1792 , 876.8208 )

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