It has been claimed from previous studies that the average diameter of ball bearings from this manufacturing process is 2.30 cm. Based on the sample of 50 that you collected, is there evidence to suggest that the average diameter is greater than 2.30 cm? Perform a hypothesis test for the population mean at alpha = 0.01. a. Define the null and alternative hypotheses in mathematical terms as well as in words.
b. Identify the level of significance.
c. Include the test statistic and the P-value.
d. Provide your conclusion and interpretation of the results. Should the null hypothesis be rejected? Why or why not?
Diameters data frame of the first sample (showing only the first five observations)
diameters
0 3.46
1 2.64
2 1.89
3 2.56
4 2.09
Diameters data frame of the second sample (showing only the first five observations)
diameters
0 3.10
1 2.04
2 2.18
3 2.60
4 2.76
test-statistic = 2.06
two tailed p-value = 0.0394

Data for all 50 samples cannot be obtained, however, the solution below uses the 10 samples below to show how the hypothesis can be tested.

Step-by-step explanation:

Average diameter, μ = 2.30

H0 : Average diameter is equal to 2.30cm

H1 : Average diameter is greater than 2.30 cm

The hypothesis :

H0 : μ = 2.30

H1 : μ > 2.30

Using the readings from the data above :

3.46, 2.64, 1.89, 2.56, 2.09, 3.10, 2.04, 2.18, 2.60, 2.76

Sample size, n = 10

Mean, xbar = ΣX/ n = 25.32 / 10 = 2.532

Sample standard deviation, s = 0.4973 (from calculator)

The test statistic :

(xbar - μ) ÷ (s/√(n))

T = (2.532 - 2.30) ÷ (0.4973/√(10))

T = 1.475

The Pvalue :

Degree of freedom, df = n - 1 ; 10 - 1 = 9

Pvalue(1.475, 9) = 0.087

Decision region :

Reject H0 ; If Pvalue < α;

Since 0.087 > 0.01 ; we fail to reject the Null and conclude that there is no evidence to suggest that the average diameter is greater than 2.30 cm

gmt is 4 hrs ahead of edt so just add four hours to 1: 14: 20 (minutes and seconds stay the same)

gmt is 5: 14: 20

hope this !

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