In: Statistics and Probability
The Acme Company just bought a new machine that makes markers. In a random sample of 250 markers from the new machine, they find that 235 markers work. They want to test if the true proportion of working markers from the new machine is greater than 90%. Assuming that they use the approximate methodology, find the p-value of the test.
The Acme Company just bought a new machine that makes markers. In a random sample of 250 markers from the new machine, they find that 240 markers work. They want to test if the true proportion of working markers from the new machine is greater than 90%. Which is the correct alternative hypothesis?
1. |
Ha: p = 0.90 |
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2. |
Ha: p ≤ 0.90 |
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3. |
Ha: p < 0.90 |
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4. |
Ha: p > 0.90 |
The Acme Company just bought a new machine that makes markers. They want to test if the true proportion of working markers from the new machine is greater than 90% at the 5% significance level. They take a random sample of 250 markers from the new machine. They find a test statistic of 2.108 with a p-value of 0.0175. Which is the correct conclusion?
1. |
There is sufficient evidence at the 5% significance level to support the claim that the sample proportion of working markers from the new machine is less than 90%. |
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2. |
There is sufficient evidence at the 5% significance level to support the claim that the proportion of all working markers from the new machine is more than 90%. |
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3. |
There is not sufficient evidence at the 5% significance level to support the claim that the sample proportion of working markers from the new machine is more than 90%. |
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4. |
There is sufficient evidence at the 5% significance level to support the claim that the sample proportion of working markers from the new machine is more than 90%. |
A golf instructor is trying out a new technique for teaching her students. She randomly selects twenty of her students. She records the score of each student before the new technique and after the new technique. Is there evidence at α = 0.05 that true mean scores before and after the technique are different? Justify fully! You can assume normal populations where needed. She does not know which test to run, so she runs two. The outputs are given. Which is the correct design and conclusion?
Matched Pairs Design Welch’s (Independent) T Test
Test Statistic = 2.268 Test Statistic = 1.218
p-value = 0.035 p-value = 0.231
1. |
Using the Welch’s design, there is sufficient evidence at the 5% significance level to support the claim that the true before and after means are different. |
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2. |
Using the Welch’s design, there is not sufficient evidence at the 5% significance level to support the claim that the true before and after means are different. |
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3. |
Using the matched pairs design, there is not sufficient evidence at the 5% significance level to support the claim that the true before and after means are different. |
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4. |
Using the matched pairs design, there is sufficient evidence at the 5% significance level to support the claim that the true before and after means are different. |
From provided information,
The correct alternative hypothesis?
----> 4.Ha : P > 0.90
They find a test statistic of 2.108 with a p-value of 0.0175. Which is the correct conclusion?
------> 2.There is sufficient evidence at the 5% significance level to support the claim that the proportion of all working markers from the new machine is more than 90%.
A golf instructor is trying out a new technique for teaching her students. She randomly selects twenty of her students. She records the score of each student before the new technique and after the new technique. Is there evidence at α = 0.05 that true mean scores before and after the technique are different? Justify fully! You can assume normal populations where needed. She does not know which test to run, so she runs two. The outputs are given. Which is the correct design and conclusion?
-------> 4.Using the matched pairs design, there is sufficient evidence at the 5% significance level to support the claim that the true before and after means are different.
Because p- value = 0.035 is less than 0.05 level of significance.
Thank You !