Answer:
0.0139
Explanation:
Given that:
The number of sample (n) = 21
The sample distribution has mean (μ) and a standard deviation of σ/√n
The z score is given as (x - mean)/ standard deviation
x = 94.8 wpm, let us assume that σ = 10 and μ = 90
Therefore: z = (x - μ) / (σ/√n) = (94.8 - 90) / (10/√21) = 2.2
To calculate the probability using Z table:
P(X>94.8) = P(Z>94.8) = 1 - P(Z<94.8) = 1 - 0.9861 = 0.0139
The probability is low that is less than 0.05, the program is more effective than the old one.
Additional information:
This question is part of a much longer question, and this would be part D which refers to a new reading program. The difference with the old data collected is that before only 10 students were tested. Now the question asks why would there be a large variation in the sample mean, from 88 words per minute to 94.8 wpm.
The first data collected had a mean of 88 wpm and a standard deviation of 10 wpm.
Answer:
The program is effective since the probability of obtaining 94.8 wpm from a larger sample doesn't fall within the normal distribution and 95% confidence interval.
Explanation:
we are given the mean = 94.8
we must calculate z = (94.8 - 88) / (10 / √21) = 6 / 2.182 = 2.7495
now we find the probability that X is between the confidence interval using a normal distribution:
P(X > 94.8) = P(z > 2.7495) = 1 - P(z < 2.7495)*
= 1 - 0.997016 = 0.002984
Since the probability of z < 2.7495 at P <0.05 is very small (only 0.002984), then we can conclude that the new method was a success.
*I used a P-value calculator using z = 2.7495 and 5% significance.
