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A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 432 gram setting. It is believed that the machine is underfilling the bags. A 29 bag sample had a mean of 425 grams with a standard deviation of 26. Assume the population is normally distributed. A level of significance of 0.02 will be used. Find the P-value of the test statistic. You may write the P-value as a range using interval notation, or as a decimal value rounded to four decimal places.

Sagot :

Answer:

-2.154

Step-by-step explanation:

We are given;

Population mean; μ = 432

Sample mean; x¯ = 425

Sample size; n = 29

standard deviation; σ = 26

Hypothesis is defined as;

Null hypothesis; H0: μ ≥ 432

Alternative hypothesis: Ha: μ < 432

Formula for the test statistic since sample size is less than 30 is;

t = (x¯ - μ)/(σ/√n)

Plugging in the relevant values;

t = (425 - 432)/(26/√29)

t = -1.45

Now,we are given significance level of 0.02; our DF = n - 1 = 29 - 1 = 28

From t-table attached, we can see that at DF of 28 and significance value of 0.02, the critical value is 2.154

However, this is a left tailed test as the rejection region is to the left.

Thus, the critical value of a left tailed test is negative. Thus, in this case our critical value will be -2.154

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