A random sample of 49 drinks from a beverage-dispensing machine gave an average
ID: 3259732 • Letter: A
Question
A random sample of 49 drinks from a beverage-dispensing machine gave an average of 12.2 ounces with a standard deviation of 0.7 ounces. At a 5% level of significance can you conclude that the true mean beverage dispensed by the machine is more than 12 ounces? (a) How large a sample is needed to obtain a 95% confidence interval for the true mean amount dispensed by the machine within an error bound of 0.1? (b) Can you conclude that the true mean value dispensed by the machine is more than 12 ounces? Use = 0.01
Explanation / Answer
Answer to the question)
n = 49
x bar = 12.2
s = 0.7
alpha = 0.05 ( or 5%)
Claim: M > 12
.
Hypothesis:
Null Hypothesis: Ho: M = 12
Alternate Hypothesis: Ha: M > 12
[Right tailed test]
.
Formula of Test Statistic is:
t = (x bar - M) / (s /sqrt(n))
t = (12.2 -12) / (0.7 /sqrt(49))
t = 2
.
df = n-1
df = 49-1
df = 48
.
In order to find the P value we refer to the T table for t=2 , df = 48 , right tailed test
[since this large df value is not provided in the table ,we can either make use of Ti83/84 calc, or excel]
The excel command to be used to find the P value is as follows:
P value = 0.0256
.
Inference:
Since the P value 0.0256 < alpha 0.05 , we reject the null hypothesis
.
Conclusion: Since the P value is less than alpha , we conclude that the mean is more than 12. Thus the claim is statistically significant.
.
Answer to part a)
Sample size for 95% confidence level , e = 0.1 , s = 0.7
z vlaue for 95% confidence level is 1.96
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Formula of sample size:
n = (z*s /e)
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On plugging the values we get:
n = (1.96 * 0.7 / 0.1)^2
n = 188.2384 ~ 189
Thus the sample size must be of 189 units
.
Answer to part b)
We already got the t = 2, with p value = 0.0256
At alpha = 0.01 , we get
inference: Since the P value 0.0256 > alpha 0.01 , we fail to reject the null
Conclusion: As the P value > alpha , we conclude that the claim that the mean is greater than 12 is statistically insignificant.
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