James says that 12 is a common factor of 3 and 4. June says 12 is a common multiple of 3 and 4. Who is correct
1 answer:
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Answer:
(a) <em>H</em>₀: <em>μ ≥ 5.70 </em><em>vs. </em><em>Hₐ</em>:<em>μ < 5.70</em>
(b) The rejection region is (<em>t₀.₀₁,₉</em> <em>≤ -2.821</em>).
(c) The value of the test statistic is -4.33.
(d) The true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa.
Step-by-step explanation:
A hypothesis test should be conducted to determine that the if the true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa.
(a)
The hypothesis is:
<em>H</em>₀: The true mean breaking strength of the new bonding adhesive is not less than 5.70 Mpa, i.e. <em>μ ≥ 5.70</em><em>.</em>
<em>Hₐ</em>: The true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa, i.e. <em>μ < 5.70</em><em>.</em>
(b)
The alternate hypothesis indicates that the hypothesis test is left-tailed.
The rejection region for the left tailed test will be towards the lower tail of the t<em>-</em>distribution curve.
The significance level of the test is: <em>α</em> = 0.01.
The critical value is:
Use the <em>t-</em>table for the critical value.
Since rejection region is in the lower tail the critical value will be negative.
Thus, the rejection region is (<em>t₀.₀₁,₉</em> <em>≤ -2.821</em>).
(c)
The test statistic value is:
Given:
Compute the value of the <em>t</em>-statistic as follows:
The value of the test statistic is -4.33.
(d)
The value of the test is less than the critical value.
This implies that the test statistic lies in the rejection region.
Hence the null hypothesis will be rejected at 1% significance level.
<u>Conclusion:</u>
As the null hypothesis is rejected it can be concluded that the true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa.
(e)
The conditions required for the <em>t-</em>test for single mean to be valid is:
- The data should be continuous.
- The parent population should be normally distributed.
- The sample should be randomly selected.