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Suppose you have purchased a filling machine for candy bags that is supposed to fill each bag with 16 oz of candy. Assume that the weights of filled bags are approximately normally distributed. A random sample of 10 bags yields the following data (in oz):

15.87, 16.02, 15.78, 15.83, 15.69, 15.81, 16.04, 15.81, 15.92, 16.10

On the basis of this data, can you conclude that the mean fill weight is actually less than 16 oz?
a. State the appropriate null and alternate hypotheses.
b. Compute the value of the test statistic.
c. Find the P-value and state your conclusion.

Sagot :

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Answer:

H0: μ ≥ 16

H1: < 16

−2.748749;

0.011275;

Reject the Null

Step-by-step explanation:

Given the data:

15.87, 16.02, 15.78, 15.83, 15.69, 15.81, 16.04, 15.81, 15.92, 16.10

Null hypothesis ; H0: μ ≥ 16

Alternative hypothesis ; H1: < 16

Sample size, n = 10

From the data:

Using calculator,

Sample mean, m = 15.887

Standard deviation, s = 0.13

The test statistic, T

(m - μ) / s/sqrt(n)

(15.887 - 16) / (0.13/sqrt(10))

= −2.748749

Using the p value from test statistic calculator :

Degree of freedom (df) = 10 - 1 = 9 at 0.05 significance level is 0.011275

Since the p value is < 0.05

0.011 < 0.05

We reject the Null and conclude that the mean fill weight is less than 16 oz

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