In: Statistics and Probability
The Whitcomb Company manufactures a metal ring for industrial engines that usually weighs about 50 ounces. A random sample of 50 of these metal rings produced the following wights (in ounces).
51 | 53 | 56 | 50 | 44 | 47 |
53 | 53 | 42 | 57 | 46 | 55 |
41 | 44 | 52 | 56 | 50 | 57 |
44 | 46 | 41 | 52 | 69 | 53 |
57 | 51 | 54 | 63 | 42 | 47 |
47 | 52 | 53 | 46 | 36 | 58 |
51 | 38 | 49 | 50 | 62 | 39 |
44 | 55 | 43 | 52 | 43 | 42 |
57 | 49 |
(a) Construct a 95% confidence interval for the average weight of the metal rings made by Whitcomb Company.
(b) Test at 5% significance level to determine if the metal rings made by Whitcomb Company have an average weight less than 50 ounces.
(Hint: you have to first calculate the sample size, sample mean, and sample standard deviation; and then apply the appropriate statistical inference techniques)
a)
Level of Significance , α =
0.05
degree of freedom= DF=n-1= 49
't value=' tα/2= 2.0096 [Excel
formula =t.inv(α/2,df) ]
Standard Error , SE = s/√n = 6.8044 /
√ 50 = 0.962293
margin of error , E=t*SE = 2.0096
* 0.96229 = 1.933801
confidence interval is
Interval Lower Limit = x̅ - E = 49.84
- 1.933801 = 47.906199
Interval Upper Limit = x̅ + E = 49.84
- 1.933801 = 51.773801
95% confidence interval is (
47.91 < µ < 51.77
)
b)
Ho : µ = 50
Ha : µ < 50
(Left tail test)
Level of Significance , α =
0.05
sample std dev , s = √(Σ(X- x̅ )²/(n-1) )
= 6.8044
Sample Size , n = 50
Sample Mean, x̅ = ΣX/n =
49.8400
degree of freedom= DF=n-1= 49
Standard Error , SE = s/√n = 6.8044 / √
50 = 0.9623
t-test statistic= (x̅ - µ )/SE = ( 49.840
- 50 ) / 0.9623
= -0.17
p-Value = 0.4343 [Excel formula
=t.dist(t-stat,df) ]
Decision: p-value>α, Do not reject null
hypothesis
Conclusion: There is not enough evidence that metal rings made by
Whitcomb Company have an average weight less than 50 ounces
Thanks in advance!
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