Math / Data & Statistics

Question2. The paralympic committee of a sitting volleyball club has indicated that the mean score achieved by the sports' members in the past was 85.9. A group of members believes that recent changes to the sitting volleyball court have led to a change in the mean score achieved by the club's members and decides to investigate this belief. A random sample of the scores, $x$ , of 100 club members was taken and is summarized by $\sum x=8350 \quad \text { and } \quad \sum(x-\bar{x})^{2}=15321$ where $\bar{x}$ denotes the sample mean. Test, at the $5 \%$ level of significance, the group's belief that the mean score of 85.9 has changed.

Studdy Solution

STEP 1

1. The sample size is $n = 100$ .
2. The population mean under the null hypothesis is $\mu_0 = 85.9$ .
3. The sample mean is denoted by $\bar{x}$ .
4. The sample variance is given by $\sum(x-\bar{x})^2 = 15321$ .
5. We are conducting a two-tailed hypothesis test at a $5\%$ significance level.

STEP 2

1. Calculate the sample mean.
2. Calculate the sample standard deviation.
3. Formulate the null and alternative hypotheses.
4. Calculate the test statistic.
5. Determine the critical value(s) and make a decision.

STEP 3

Calculate the sample mean $\bar{x}$ :
$\bar{x} = \frac{\sum x}{n} = \frac{8350}{100} = 83.5$

STEP 4

Calculate the sample standard deviation $s$ :
$s = \sqrt{\frac{\sum(x-\bar{x})^2}{n-1}} = \sqrt{\frac{15321}{99}}$
$s \approx \sqrt{154.7576} \approx 12.44$

STEP 5

Formulate the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): $\mu = 85.9$ - Alternative hypothesis ( $H_a$ ): $\mu \neq 85.9$

STEP 6

Calculate the test statistic $t$ :
$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{83.5 - 85.9}{12.44/\sqrt{100}}$
$t = \frac{-2.4}{1.244} \approx -1.93$

STEP 7

Determine the critical value(s) for a two-tailed test at $\alpha = 0.05$ with $df = 99$ :
- Critical values for $t$ at $\alpha = 0.05$ (two-tailed) are approximately $\pm 1.984$ .
Compare the calculated $t$ -value to the critical values:
Since $-1.93$ is not less than $-1.984$ and not greater than $1.984$ , we fail to reject the null hypothesis.
Conclusion: There is not enough evidence at the $5\%$ level of significance to conclude that the mean score has changed from 85.9.

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Studdy Solution

STEP 1

1. The sample size is $n = 100$ .
2. The population mean under the null hypothesis is $\mu_0 = 85.9$ .
3. The sample mean is denoted by $\bar{x}$ .
4. The sample variance is given by $\sum(x-\bar{x})^2 = 15321$ .
5. We are conducting a two-tailed hypothesis test at a $5\%$ significance level.

STEP 2

1. Calculate the sample mean.
2. Calculate the sample standard deviation.
3. Formulate the null and alternative hypotheses.
4. Calculate the test statistic.
5. Determine the critical value(s) and make a decision.

STEP 3

Calculate the sample mean $\bar{x}$ :
$\bar{x} = \frac{\sum x}{n} = \frac{8350}{100} = 83.5$

STEP 4

Calculate the sample standard deviation $s$ :
$s = \sqrt{\frac{\sum(x-\bar{x})^2}{n-1}} = \sqrt{\frac{15321}{99}}$
$s \approx \sqrt{154.7576} \approx 12.44$

STEP 5

Formulate the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): $\mu = 85.9$ - Alternative hypothesis ( $H_a$ ): $\mu \neq 85.9$

STEP 6

Calculate the test statistic $t$ :
$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{83.5 - 85.9}{12.44/\sqrt{100}}$
$t = \frac{-2.4}{1.244} \approx -1.93$

STEP 7

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Studdy Solution

STEP 1

STEP 2

1. Calculate the sample mean.2. Calculate the sample standard deviation.3. Formulate the null and alternative hypotheses.4. Calculate the test statistic.5. Determine the critical value(s) and make a decision.

STEP 3

Calculate the sample mean xˉ \bar{x} xˉ:xˉ=∑xn=8350100=83.5 \bar{x} = \frac{\sum x}{n} = \frac{8350}{100} = 83.5 xˉ=n∑x​=1008350​=83.5

STEP 4

Calculate the sample standard deviation s s s:s=∑(x−xˉ)2n−1=1532199 s = \sqrt{\frac{\sum(x-\bar{x})^2}{n-1}} = \sqrt{\frac{15321}{99}} s=n−1∑(x−xˉ)2​​=9915321​​s≈154.7576≈12.44 s \approx \sqrt{154.7576} \approx 12.44 s≈154.7576​≈12.44

STEP 5

Formulate the null and alternative hypotheses:- Null hypothesis (H0 H_0 H0​): μ=85.9 \mu = 85.9 μ=85.9 - Alternative hypothesis (Ha H_a Ha​): μ≠85.9 \mu \neq 85.9 μ=85.9

STEP 6

Calculate the test statistic t t t:t=xˉ−μ0s/n=83.5−85.912.44/100 t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{83.5 - 85.9}{12.44/\sqrt{100}} t=s/n​xˉ−μ0​​=12.44/100​83.5−85.9​t=−2.41.244≈−1.93 t = \frac{-2.4}{1.244} \approx -1.93 t=1.244−2.4​≈−1.93

STEP 7

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1. Calculate the sample mean.
2. Calculate the sample standard deviation.
3. Formulate the null and alternative hypotheses.
4. Calculate the test statistic.
5. Determine the critical value(s) and make a decision.

Calculate the sample mean $\bar{x}$ :
$\bar{x} = \frac{\sum x}{n} = \frac{8350}{100} = 83.5$

Calculate the sample standard deviation $s$ :
$s = \sqrt{\frac{\sum(x-\bar{x})^2}{n-1}} = \sqrt{\frac{15321}{99}}$
$s \approx \sqrt{154.7576} \approx 12.44$

Formulate the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): $\mu = 85.9$ - Alternative hypothesis ( $H_a$ ): $\mu \neq 85.9$

Calculate the test statistic $t$ :
$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{83.5 - 85.9}{12.44/\sqrt{100}}$
$t = \frac{-2.4}{1.244} \approx -1.93$