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A manufacturer of giftbox claims that the giftboxes sold weigh on average at least 16 ounces. The distribution of weight is known to be normal, with a standard deviation of 0.4 ounce. A random sample of 16 giftboxes yielded a sample mean weight of 15.84 ounces. Test at the 10% significance level the null hypothesis that the population mean weight is 16 ounces. Specifically, follow the steps below.

a. State the null hypothesis.
b. State the alternative hypothesis.
c. State the level of significance.
d. Determine the appropriate test statistic and the critical values that divide the rejection and
nonrejection regions.
e. Compute the value of the test statistic.
f. Make the statistical decision.
g. Compute the p-value and make the statistical decision according to p-value.

Respuesta :

Chrisnando

Answer:

a) Null hypothesis:

H0 : u = 16

b) Alternative hypothesis :

H1 : u < 16

c) Level of significance, a = 10% = 0.10

d) To find the test statistic Z, let's use the formula below:

[tex] Z = \frac{x' - u}{\sigma/ \sqrt{n}} [/tex]

Where,

x' = 15.84

u = 16

[tex] \sigma = 0.4 [/tex]

n = 16

[tex] Z = \frac{15.84 - 16}{0.4/ \sqrt{16}} [/tex]

Z = -1.6

This is a left tailed test

Critical value of test statistic at significance level of [tex]Z_0_._1_0 = -1.282[/tex]

f) Decision: reject null hypothesis H0 if Z is less than critical value.

g) From standard normal table, pvalue at Z = - 1.6

Pvalue = 0.0548

Conclusion: Since pvalue,0.0548 is less than significance level of 0.10  and Zo, - 1.6 is less than Zcritical, -1.282 we reject the null hypothesis H0.

Therefore, we conclude that the population mean weight is less than 16oz.

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