A pharmaceutical manufacturer forms tablets by compressing a granular material that contains the active ingredient and var. fillers. The hardness of a sample from each lot of tablets produced is measured to control compression. The target values for the hardness are mean=11.5, standard dev=0.2. The hardness data for sample of 20 tablets:
11.627, 11.613, 11.493, 11.602, 11.630, 11.374, 11.592, 11.458, 11.552, 11.463, 11.383, 11.715, 11.485, 11.509, 11.429, 11.477, 11.570, 11.623, 11.472, 11.531
Is there signifcant evidence at the 5% level that the mean hardness of tablets is different from the target value?
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Okay, I actually solved most of the problem up to the part where you have to find the P-value. I have no idea how to do it....like how do you know when to multiply by 2P? Also, using Ti-83 calculator, how do you use normalcdf function to find out the percentage that you need to interpret the results to see whether it rejects the null hypothesis? What do you put into the normal cdf function to get that percentage? Please help asap thanks
If you are doing a one-sample t-test, then the value you found for the test statistic can be used to find your p-value. The p-value is the actual level of the test statistic. It represents the probability of getting a result as extreme as the test statistic itself. Also, if you are just looking for a significant difference at the 5% level, you can check a t-table for the cutoff points at both ends of the distribution (divide 5% in half for a two-tailed test). If your test statistic exceeds the positive or negative critical value (or cutoff point), then the null is rejected in favor of the alternate or alternative hypothesis. If your test statistic does not exceed the positive or negative critical value, then the null is not rejected.
Someone else may be able to help you with the calculator functions.
To find the p-value in this case, you can use the t-distribution.
First, calculate the test statistic, which is the t-value. The formula for the t-value is:
t = (sample mean - target mean) / (sample standard deviation / sqrt(sample size))
In this case, the target mean is 11.5, the sample mean will be the mean of the hardness values you provided, the sample standard deviation will be the standard deviation of the hardness values you provided, and the sample size is 20.
Next, you can use the t-distribution table or a calculator to find the p-value. The p-value is the probability of getting a test statistic more extreme than the one you calculated, assuming the null hypothesis is true.
If you are using a TI-83 calculator, you can use the function "tcdf" (t cumulative distribution function) to find the p-value. The "tcdf" function takes three inputs:
tcdf(lower limit, upper limit, degrees of freedom)
To find the percentage to interpret the results and see whether it rejects the null hypothesis, you will need to calculate the p-value and compare it to the significance level of 5% (0.05).
If the p-value is less than the significance level, it means the test statistic is unlikely to occur by chance alone, so you would reject the null hypothesis and conclude that there is significant evidence that the mean hardness of the tablets is different from the target value. If the p-value is greater than the significance level, it means the test statistic is likely to occur by chance alone, so you would fail to reject the null hypothesis and conclude that there is not enough evidence to suggest a significant difference.
I hope this clarifies the process for finding the p-value and interpreting the results.
To find the p-value using a calculator like the TI-83, you can use the `normalcdf` function. The `normalcdf` function calculates the cumulative probability between two values for a normal distribution.
In this case, you want to find the probability of getting a test statistic as extreme as the one you calculated. Since this is a two-tailed test, you need to find the probability in both tails.
To use `normalcdf` on the TI-83, follow these steps:
1. Press the "2nd" (second) button, then press "VARS" (DISTR).
2. Scroll down to "2:normalcdf(" and press "Enter."
3. Enter the lower and upper bounds for the left tail probability. For a two-tailed test, you can use the absolute value of your test statistic as the upper bound for the left tail. Press "Enter."
4. Enter the upper and lower bounds for the right tail probability. Again, use the absolute value of your test statistic as the lower bound for the right tail. Press "Enter."
5. The result displayed will be the combined probability for both tails.
By comparing the p-value to the significance level (0.05 in this case), you can determine if there is significant evidence to reject the null hypothesis.
If the p-value is less than the significance level (p-value < 0.05), you would reject the null hypothesis. If the p-value is greater than or equal to the significance level (p-value >=0.05), you would fail to reject the null hypothesis.
I hope this helps! Let me know if you have any other questions.