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9.27 In contract negotiations a company claims that a new incentive scheme has resulted in average weekly earnings of at least \$ 400 for all customer service workers. A union representative takes a random sample of 15 workers and finds that their weekly earnings have an average of \$ 381.35 and a standard deviation of \$ 48.60. Assume a normal distribution. a. Test the company’s claim. b. If the same sample results had been obtained from a random sample of 50 employees, could the company’s claim be rejected at a lower significance level than that used in part a?
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From the Z-Test link We could also say that with 98.6% confidence we reject the null hypothesis that the 55 test takers are comparable to a simple random sample from the population of test-takers. Click to expand... So that looks like it translates to a null hypothesis that 400 is comparable to a measured random sample. For the 15-sample measure, we could only reject the null hypothesis with a confidence of 86.38%, which isn't really good enough to reject it. For the 50-sample measure, we could reject the null hypothesis with a confidence of 99.33% which is generally considered statistically significant. And I should be able to state any preferred threshold point by rearranging the equation $z = \frac{x - \mu}{\sigma_{\overline{x}}}$ to solve for x $x = z \times \sigma_{\overline{x}} + \mu = 1.96 \times 12.55 + 381.35 \simeq 406$ for the 95% rejection threshold in the 15-sample case, and $1.96 \times 6.87 + 381.35 \simeq 395$ for the 95% rejection threshold in the 50-sample case.
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AngleWyrm said: So if I'm understanding that right, Standard Error for 15 samples: $=\frac{SD(X_{i})}{\sqrt{n}}=\frac{48.60}{\sqrt{15}} \simeq$ 12.55 Standard Error for 50 samples: $=\frac{SD(X_{i})}{\sqrt{n}}=\frac{48.60}{\sqrt{50}} \simeq$ 6.87 So then a z-score based on Standard Error would be $z_{15} = \frac{\text{observation - mean}}{\text{standard error}} = \frac{400-381.35}{12.55} \simeq$ 1.49 $z_{50}= \frac{\text{observation - mean}}{\text{standard error}} = \frac{400-381.35}{6.87} \simeq$ 2.71 Then this calculator describes the z scores in terms of percentage areas My first guess: z=1.49 confidence: 86.38%, two-tail region: 13.62% z=2.71 confidence: 99.33%, two-tail region: 0.67% So is it reasonable then to say with a sample size of 15 there's a 13.62% assurance the value 400 is drawn from the same population, but with a sample size of 50 that assurance drops to 0.67%? Or is this a one-tail problem? Also, when is it the right thing to do to use Standard Error vs Standard Deviation? Click to expand... Depending on which test you're performing, the test statistic is different. In this particular exercise, we're performing a Z-test, which uses the standard error.
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AngleWyrm said: So if I'm understanding that right, Standard Error for 15 samples: $=\frac{SD(X_{i})}{\sqrt{n}}=\frac{48.60}{\sqrt{15}} \simeq$ 12.55 Standard Error for 50 samples: $=\frac{SD(X_{i})}{\sqrt{n}}=\frac{48.60}{\sqrt{50}} \simeq$ 6.87 So then a z-score based on Standard Error would be $z_{15} = \frac{\text{observation - mean}}{\text{standard error}} = \frac{400-381.35}{12.55} \simeq$ 1.49 $z_{50}= \frac{\text{observation - mean}}{\text{standard error}} = \frac{400-381.35}{6.87} \simeq$ 2.71 Then this calculator describes the z scores in terms of percentage areas My first guess: z=1.49 confidence: 86.38%, two-tail region: 13.62% z=2.71 confidence: 99.33%, two-tail region: 0.67% So is it reasonable then to say with a sample size of 15 there's a 13.62% assurance the value 400 is drawn drawn from the same population, but with a sample size of 50 that assurance drops to 0.67%? Or is this a one-tail problem? Also, when is it the right thing to do to use Standard Error vs Standard Deviation? Click to expand... Strictly there are no fixed rules other than if you are discussing a sample mean the SE is appropriate If you are discussing a proportion of a population use the SD. But the final criteria is: Think what you are trying to do and select an appropriate approach. Virtually every real statistical problem is different and requires a method appropriate to itself, not a potted test/method.
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So if I'm understanding that right, Standard Error for 15 samples: $=\frac{SD(X_{i})}{\sqrt{n}}=\frac{48.60}{\sqrt{15}} \simeq$ 12.55 Standard Error for 50 samples: $=\frac{SD(X_{i})}{\sqrt{n}}=\frac{48.60}{\sqrt{50}} \simeq$ 6.87 So then a z-score based on Standard Error would be $z_{15} = \frac{\text{observation - mean}}{\text{standard error}} = \frac{400-381.35}{12.55} \simeq$ 1.49 $z_{50}= \frac{\text{observation - mean}}{\text{standard error}} = \frac{400-381.35}{6.87} \simeq$ 2.71 Then this calculator describes the z scores in terms of percentage areas My first guess: z=1.49 confidence: 86.38%, two-tail region: 13.62% z=2.71 confidence: 99.33%, two-tail region: 0.67% So is it reasonable then to say with a sample size of 15 there's a 13.62% assurance the value 400 is drawn drawn from the same population, but with a sample size of 50 that assurance drops to 0.67%? Or is this a one-tail problem? Also, when is it the right thing to do to use Standard Error vs Standard Deviation?
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AngleWyrm said: Could you please go into some more detail on those results? My understanding of a z-score is $z = \frac{\text{observation - mean}}{\text{standard deviation}} = \frac{400 - 381.35}{48.60} \simeq$ 0.3837 So about 0.38 standard deviations from mean $0.3837 \times 48.60 =$ 18.65 $18.65 + 381.35 =$ 400 Click to expand... As @CaptainBlack stated, you'd want to look at the standard error, and not the standard deviation of 1 sample. i.e. $Var(\bar{X})$ and not $Var(X_i)$. Assume i.i.d: $$Var(\bar{X})=Var\left(\frac{\sum_{i=1}^{n}X_i}{n}\right)=\frac{1}{n^2}Var\left(\sum_{i=1}^{n}X_i\right)=\frac{1}{n^2}\cdot n\cdot Var(X_i)=\frac{Var(X_i)}{n}$$ Taking the square root, the standard error is: $$\frac{SD(X_i)}{\sqrt{n}}$$ Notice as $n$ increases, the standard error decreases. So, the sample size is something to concern about.
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Could you please go into some more detail on those results? My understanding of a z-score is $z = \frac{\text{observation - mean}}{\text{standard deviation}} = \frac{400 - 381.35}{48.60} \simeq$ 0.3837 So about 0.38 standard deviations larger than mean $0.3837 \times 48.60 =$ 18.65 $18.65 + 381.35 =$ 400
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AngleWyrm said: 400 is well within 1 standard deviation (48.60) of 381.35, so the claim is not far off. Click to expand... You should be looking at the standard error of the sample mean not the standard deviation of the sample. Even so the z-score of the difference of the sample mean from the claim is ~1.5 so it is not inconsistent with the claim. If the sample size had been 50 not 15 then the z-score would heve been ~2.7, and so something to worry about.
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400 is well within 1 standard deviation (48.60) of 381.35, so the claim is not far off.
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