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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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