A teacher instituted a new reading program at school. after 10 weeks in the program, it was found that the mean reading speed of a random sample of 21 second grade students was 94.8 wpm. what might you conclude based on this result? select the correct choice below and fill in the answer boxes within your choice

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.