Mr. Rodriguez is packing bags of snacks for his children’s lunchboxes. If he has 20 blueberries and 30 grapes, what is the maxim
2 answers:
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Answer:
(a) There is enough evidence to support the claim that supplier 2 provides gears with higher mean impact strength.
(b) <em>p</em>-value = 0.033.
(c) The 95% confidence interval estimate for the difference in mean impact strength between supplier 2 and supplier 1 is (-64.26, 2.26).
(d) The null hypothesis is rejected.
Step-by-step explanation:
Let <em>X</em>₁ denotes plastic gear manufactured by supplier 1 and <em>X</em>₂ denotes plastic gear manufactured by supplier 2.
The given information is,
(a)
The hypothesis for the test can be defined as:
<em>H₀</em>: There is no difference between the mean impact strength of the gears provided by the two suppliers, i.e. <em>μ</em>₁ - <em>μ</em>₂ = 0.
<em>Hₐ</em>: The means impact strength of the gears provided by the supplier 2 is higher, i.e. <em>μ</em>₁ - <em>μ</em>₂ < 0.
It is assumed that the two populations are normally distributed and the variances are equal.
We will use a <em>t</em>-test to perform the test.
The t-statistic is given by,
= pooled standard deviation
Compute the pooled standard deviation as follows:
Compute the test statistic as follows:
The, <em>t</em>-statistic value is -1.92.
The degrees of freedom of the test is:
<em>df</em> = (n₁ + n₂ - 2) = 24
Decision rule:
If the test statistic value is less than the critical value then the null hypothesis will rejected.
The critical value is:
*Use a <em>t</em>-table.
The test statistic value is less than the critical value.
Thus, the null hypothesis will be rejected at 5% level of significance.
So, there is enough evidence to support the claim that supplier 2 provides gears with higher mean impact strength.
(b)
For the computed <em>t</em>-statistic and (n₁ + n₂ - 2) degrees of freedom, the <em>p</em>-value will be,
Use the <em>t</em>-table.
The <em>p</em>-value of the test is less than the significance level . Thus, the null hypothesis is rejected.
Concluding that there is enough evidence to support the claim that supplier 2 provides gears with higher mean impact strength.
(c)
The 95% confidence interval is:
Thus, the 95% confidence interval estimate for the difference in mean impact strength between supplier 2 and supplier 1 is (-64.26, 2.26).
(d)
A confidence interval can be used to test the claim made.
If the confidence interval consist the null value of the parameter then the null hypothesis will be accepted or else rejected.
The alternate hypothesis to be tested is:
<em>Hₐ</em>: The mean impact strength of gears from supplier 2 is at least 25 foot-pounds higher than that of supplier 1, i.e. <em>μ</em>₁ - <em>μ</em>₂ ≥ - 25
The 95% confidence interval estimate for the difference in mean impact strength consist the difference values less than 25 foot-pounds.
Thus, the null hypothesis is rejected.