Question

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

Answer 1

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.

Answer 2

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.