Instead, you should be using the t distribution with 39 degrees of freedom.

In summary, the conversation discusses finding the upper 95% confidence interval and testing a null hypothesis for a measure of contamination with unknown mean and variance. The sample mean and variance are given as 28.30 and 17.38 respectively. The attempt at a solution uses the formula for the confidence interval and the z-score for the test, but these may not be applicable given the unknown variance.
  • #1
joemama69
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Homework Statement



Measure of Contaminatino has a normal distribution w/ unknown mean and unknown variance. Random sample of n=40 provides sample mean = 28.30 and sample variance = 17.38

1)Find Upper 95% Confidence Interval

2)Test Null Hyp = 25.5 vs Alt Hyp = 28.30... Upper Tailed

Homework Equations





The Attempt at a Solution



1)...95% CI... therefore z(1.645) = 95%

(xbar)+z sqrt(sigma^2 / n) = 28.30 + 1.645 sqrt(17.38 / 40) = 28.30 + 1.0843=29.38

(0 < u < 29.38)


2) Test Null Hyp = 25.5 vs Alt Hyp = 28.30... Upper Tailed

Since n = 40 we do not to use T-Tables and can just compute the P-Value from Z-Table.

Z=(x-u)/sqrt(sigma^2/n) = (28.30 - 25.50)/sqrt(17.38 / 40) = z(4.25) = 1.00 ?

Since it us upper tailed P-Value = 1-z(4.25) = 1-1 = 0 ?


Where did I go wrong... Based on my 95% CI, Shouldnt I get a little less than 95% for a value lessthan 29.38?
 
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  • #2
joemama69 said:

Homework Statement



Measure of Contaminatino has a normal distribution w/ unknown mean and unknown variance. Random sample of n=40 provides sample mean = 28.30 and sample variance = 17.38

1)Find Upper 95% Confidence Interval

2)Test Null Hyp = 25.5 vs Alt Hyp = 28.30... Upper Tailed

Homework Equations





The Attempt at a Solution



1)...95% CI... therefore z(1.645) = 95%

(xbar)+z sqrt(sigma^2 / n) = 28.30 + 1.645 sqrt(17.38 / 40) = 28.30 + 1.0843=29.38

(0 < u < 29.38)


2) Test Null Hyp = 25.5 vs Alt Hyp = 28.30... Upper Tailed

Since n = 40 we do not to use T-Tables and can just compute the P-Value from Z-Table.

Z=(x-u)/sqrt(sigma^2/n) = (28.30 - 25.50)/sqrt(17.38 / 40) = z(4.25) = 1.00 ?

Since it us upper tailed P-Value = 1-z(4.25) = 1-1 = 0 ?


Where did I go wrong... Based on my 95% CI, Shouldnt I get a little less than 95% for a value lessthan 29.38?

The questions make no sense: 95% confidence interval of WHAT? The mean? The variance? The actual level of contamination?

If you do want a confidence interval for the mean, you are using the wrong distribution: when you have to estimate both the mean and the variance from the data, the normal distribution does not apply. (It would apply if you knew the variance exactly and only had to estimate the mean from the data.)
 

1. What is a P-Value for Unknown Variance?

A P-Value for Unknown Variance is a statistical measure that is used to determine the likelihood of obtaining a result at least as extreme as the one observed in a sample, under the assumption that the null hypothesis is true and the population variance is unknown.

2. How is the P-Value for Unknown Variance calculated?

The P-Value for Unknown Variance is calculated by first determining the test statistic, which is typically the ratio of the sample variance to the hypothesized population variance. This test statistic is then compared to a critical value from a statistical table or calculated using a statistical software. The P-Value is then calculated as the probability of obtaining a test statistic at least as extreme as the one observed in the sample, assuming the null hypothesis is true.

3. What is the significance of the P-Value for Unknown Variance?

The P-Value for Unknown Variance is used to determine the statistical significance of a hypothesis test. A small P-Value (typically less than 0.05) indicates that the observed result is unlikely to occur by chance and therefore the null hypothesis can be rejected. A larger P-Value suggests that the null hypothesis cannot be rejected and the observed result may be due to chance.

4. What are some limitations of the P-Value for Unknown Variance?

One limitation of the P-Value for Unknown Variance is that it does not provide information about the effect size or the strength of the relationship between variables. Additionally, the P-Value is influenced by sample size and can be affected by outliers or non-normal data. It is important to consider these limitations when interpreting the results of a hypothesis test using the P-Value for Unknown Variance.

5. How should the P-Value for Unknown Variance be interpreted?

The P-Value for Unknown Variance should be interpreted in the context of the specific research question and the study design. It is important to carefully consider the sample size, effect size, and potential limitations when interpreting the P-Value. Additionally, the P-Value should not be the sole factor in determining the significance of a result, and should be considered alongside other statistical measures and research findings.

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