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a sample of 49 parents found the mean time spent posting pictures of their children on social media per week was 19 minutes with a standard deviation of 3.9 minutes. find the p-value used to test a claim that the mean time parents spend posting pictures of their children per week is more than 18 minutes at the 5% significance level.

Sagot :

The p-value will be 0.9457

Sample size = 49

Meantime = 19 minutes

Standard deviation = 3.9 mins

Calculating the z-score for test statistic -

z = (x - μ) / (s / √n)

z = (19 - 18) / (3.9 / √49)

= 1 / 0.63

= 1.58

To find the p-value, we can use a standard normal table to find area under the normal curve of the z-score.

The p-value will be the area under the curve to the right of 1.58.

This value is 0.9457, which means that there is a 94.57% chance.

Now,

Since the p-value is greater than the significance level (0.9457 > 0.05),  the null hypothesis can not be rejected. Therefore, the mean time parents spend posting pictures of their children per week is not significantly different from 18 minutes at the 5% significance level.

Read more about p-value on:

https://brainly.com/question/28027137

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