Question

A researcher wishes to test the claim that for a particular manufacturer of cereal, the mean weight in its boxes of cereal is less than 18 ounces. A sample of 36 boxes yields a sample mean weight of 17.88 ounces. Assume that the population standard deviation is .28 ounces. Let a =.05.

a. Conduct a 5-step test of hypotheses:

i. H0:

Ha:

ii. a=_______.

iii. Test Statistic (if a t test is used, also report the df):

iv. P-value:

v. Conclusion (both statistical conclusion and conclusion in the context of the problem) :

b. Find a 95% confidence interval for u= true mean weight.

Answer 1

Here, we have given that,

Claim: To check whether the mean weight in its boxes of cereal is less than 18 ounces.

The Hypothesis is as follows

$Ho:μ=18$

v/s

$H1: \mu < 18$

We have given that,

n= Number of observation = 36

$\bar x$ = sample mean weight =17.88 ounces

$\sigma$ = population standard deviation = 0.28

Here population standard deviation is given here we use the Z standard normal test.

Now, we can find the test statistic

$Z-statistics= \frac{\bar x- \mu}{\frac{\sigma}{\sqrt(n)}}$

$= \frac{ 17.88- 18}{\frac{0.28}{\sqrt(46)}}$

= -2.91

we get,

the Test statistic is -2.91

Now we find the P-value

$\alpha$ = level of significance=0.05

This is Left one tailed test

Now, we can find the P-value

P-value =(P(Z < z) = P( Z < -2.91)=0.0018 using standard normal z probability table

we get the P-value is 0.0018

Decision:

P-value < 0.05 ( $\alpha$ )

That is we reject Ho (Null Hypothesis)

Conclusion

There is the sufficient evidence that mean weight in its boxes of cereal is less than 18 ounces

(B)

Now, we want to find

The 95% confidence inteval for $\mu$ = true mean weight 18

Formula is as follows,

$\bar x - E \leq \mu \leq \bar x + E$

Where

E=Margin of error = $Z(critical)*\frac{\sigma}{\sqrt(n)}$

Now,

Degrees of freedom = n-1 = 36-1=35

c=confidence level =0.95

$\alpha$ =level of significance=1-c=1-0.95=0.05

and we know that confidence interval is always two tailed

Zcritical= $Z\frac{\alpha}{2}$ = $Z\frac{0.05}{2}$ = 1.96 using Excel =NORMSINV(PROB=0.05/2))

Now,

E= margin of error = $Z(critical)*\frac{\sigma}{\sqrt(n)}$ = $1.96*\frac{0.28}{\sqrt(36)}$ =0.091

We get the 95% confidence interval for the population mean $\mu$

$\bar x - E \leq \mu \leq \bar x + E$

$17.88 - 0.091 \leq \mu \leq 17.88 + 0.091$

$17.79 \leq \mu \leq 17.97$

Interpretation:

This is the 95% CI which shows that we have 95% confidence that this population mean will fall within this interval.