A random sample of size 20 from a normal population will sample mean 42 and sample standard deviation of 6 . To test the hypothesis the population mean 44 at 5% level of significance , find the p-value and comment
Ho: mean1 = mean2
H1: mean1 ≠ mean2
Z = (mean 1 -mean2)/ (SD/sq rt N)
Find Z value and look it up in table in the back of your statistics text called something like "areas under normal distribution" to get p value.
I hope this helps.
Well, well, well! It looks like our sample mean (42) is quite a distance away from the hypothesized population mean (44).
To calculate the p-value, we can use a one-sample t-test. With a sample size of 20 and a sample standard deviation of 6, we can calculate the t-value as follows:
t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))
= (42 - 44) / (6 / sqrt(20))
= -2 / 1.3416407865
= -1.49
Now, we need to determine the p-value associated with this t-value. Since the test is a one-tailed test (we're only checking if the sample mean is less than the hypothesized mean), we'll look up the p-value using the t-distribution for degrees of freedom of n-1 (20-1 = 19).
Using a t-table or calculator, the p-value is approximately 0.0808.
Finally, let's compare the p-value to the significance level (5%) to determine our conclusion. Since the p-value (0.0808) is greater than the significance level (0.05), we fail to reject the null hypothesis. In other words, we do not have enough evidence to conclude that the population mean is different from 44 at the 5% significance level.
Remember, p-values measure the strength of evidence against the null hypothesis. In this case, the evidence is just not that strong. Maybe the population mean is 44 after all, and our sample was just clowning around!
To test the hypothesis, we can use a one-sample t-test. Here are the steps to calculate the p-value:
Step 1: State the null and alternative hypotheses.
Null hypothesis: The population mean is equal to 44.
Alternative hypothesis: The population mean is not equal to 44.
Step 2: Set the level of significance.
The given significance level is 5% or 0.05.
Step 3: Calculate the test statistic.
Given that the sample size is 20, sample mean is 42, and sample standard deviation is 6, we can calculate the test statistic using the formula:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
t = (42 - 44) / (6 / sqrt(20))
Step 4: Determine the degrees of freedom.
The degrees of freedom for a one-sample t-test is simply the sample size minus 1.
Degrees of freedom = 20 - 1 = 19.
Step 5: Calculate the p-value.
Using the t-distribution table or a statistical calculator, we can determine the p-value associated with the test statistic and degrees of freedom. The p-value is the probability of observing a test statistic as extreme or more extreme than the one calculated.
Step 6: Compare the p-value to the significance level.
If the p-value is less than the significance level (0.05), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Without the exact test statistic value, we cannot calculate the p-value. However, if you provide the exact test statistic value, I can help you calculate the p-value and proceed with the analysis.
To calculate the p-value, you need to perform a hypothesis test using the given information. Here are the steps involved:
Step 1: State the null and alternative hypotheses:
- Null hypothesis (H0): The population mean is 44 (µ = 44).
- Alternative hypothesis (Ha): The population mean is not equal to 44 (µ ≠ 44).
Step 2: Determine the test statistic:
Since the sample size is greater than 30 and the population standard deviation is not known, you can use the t-test. The formula for the t-test statistic is:
t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))
Plugging in the given values:
t = (42 - 44) / (6 / sqrt(20))
Step 3: Find the degrees of freedom:
The degrees of freedom for a t-test are (sample size - 1). In this case, it is (20 - 1) = 19.
Step 4: Calculate the p-value:
Using the test statistic and degrees of freedom, you can find the p-value. Since the alternative hypothesis is two-tailed (µ ≠ 44), you need to find the probability of getting a test statistic less extreme than the observed value in both tails.
For a two-tailed test at a 5% significance level:
- The area in one tail is 0.05/2 = 0.025.
You can find the p-value using a t-distribution table or a calculator. In this case, the t-value is -1.7884 (calculated using the t-distribution table).
Step 5: Interpret the results:
If the p-value is less than the significance level (0.05), then the null hypothesis is rejected. Otherwise, you fail to reject the null hypothesis.
Since the p-value is not provided, you can interpret the results based on the critical value approach:
- If the t-value obtained is larger than the critical value at the 5% level of significance (|t| > 1.729), then you can reject the null hypothesis.
- If the t-value obtained is smaller than the critical value at the 5% level of significance (|t| < 1.729), then you fail to reject the null hypothesis.
Therefore, you need to compare the t-value (-1.7884) with the critical value (-1.729). Since the t-value is less than the critical value, the p-value will be greater than 0.05, indicating that you fail to reject the null hypothesis.