Question

1. a study finds that the mean score on a test for a sample of 64 students is 78, with a standard deviation of 6 points. calculate the margin of error (me) at a 95% confidence level (z = 1.96).

a. 1.45 points
b. 1.47 points
c. 1.50 points
d. 1.43 points

Explanation:

Step1: Recall the formula for margin of error

The formula for margin of error (ME) when the population standard - deviation is unknown (we use sample standard deviation $s$ as an estimate) and the sample size $n$ is large ($n\geq30$) is $ME = z\times\frac{s}{\sqrt{n}}$, where $z$ is the z - score, $s$ is the sample standard deviation, and $n$ is the sample size.

Step2: Identify the values of $z$, $s$, and $n$

We are given that $z = 1.96$ (for a 95% confidence level), $s = 6$ (sample standard deviation), and $n=64$ (sample size).

Step3: Calculate the margin of error

Substitute the values into the formula: $ME=1.96\times\frac{6}{\sqrt{64}}$. First, $\sqrt{64}=8$. Then $\frac{6}{8}=0.75$. Finally, $ME = 1.96\times0.75=1.47\approx1.50$ (rounded to two decimal places).

Answer:

C. 1.50 points