Explaination on p-value
In the video "Test for the mean. Population variance unknown", there is a mention of p-value as 0.304. I am not able to understand how we are getting that value.
In the previous videos we have seen that for one sided test :
p-value = 1 - critical value from z-table
But in this example we are dealing with t-table. Can you explain how we are coming up with that value of 0.304 ?
Regards,
Sayantan Mukhopadhyay
In the previous videos we have seen that for one sided test :
p-value = 1 - critical value from z-table
But in this example we are dealing with t-table. Can you explain how we are coming up with that value of 0.304 ?
Regards,
Sayantan Mukhopadhyay
1 answers ( 0 marked as helpful)
Hi Sayantan!
Thanks for reaching out.
In this example, we seek the p-value for a z-score of 2.12. Therefore, we look in the z-table for the p-value corresponding to a z score of 2.1
We first find the row for 2.1 and then the column for 0.02 in the z-table (because we use the z-score's first decimal to find the row in the table and the second decimal to locate the column). The value at this intersection is 0.9830. The p-value for a one-sided test is 1 - the value from the table, which gives us 1 - 0.9830, or 0.017. For a two-sided test, we multiply this value and obtain 0.034.
Regarding the t-score, the cumulative probability (area to the left of t=−0.53 since we're dealing with a one-sied test) in the t-distribution is approximately 0.304468. You can check this with an online statistical calculator considering a one-sided test with a significance level of 0.05 and degrees of freddom: 9. It's not directly observable in the t-table.
Consequently, since the p-value (0.304) is greater than 0.05, we fail to reject the null hypothesis.
Hope this helps.
Best,
Ivan
Thanks for reaching out.
In this example, we seek the p-value for a z-score of 2.12. Therefore, we look in the z-table for the p-value corresponding to a z score of 2.1
We first find the row for 2.1 and then the column for 0.02 in the z-table (because we use the z-score's first decimal to find the row in the table and the second decimal to locate the column). The value at this intersection is 0.9830. The p-value for a one-sided test is 1 - the value from the table, which gives us 1 - 0.9830, or 0.017. For a two-sided test, we multiply this value and obtain 0.034.
Regarding the t-score, the cumulative probability (area to the left of t=−0.53 since we're dealing with a one-sied test) in the t-distribution is approximately 0.304468. You can check this with an online statistical calculator considering a one-sided test with a significance level of 0.05 and degrees of freddom: 9. It's not directly observable in the t-table.
Consequently, since the p-value (0.304) is greater than 0.05, we fail to reject the null hypothesis.
Hope this helps.
Best,
Ivan