Math  /  Data & Statistics

QuestionC8 Q5 V3: A random sample of 100 students is taken from a larger population of students in a multi-lecture course and the sample mean (average) of their final grades is found to be 63.25. Suppose it is known that the population standard deviation is 3.4. You wish to determine whether the average grade for the population differs from 62.75 . Use R to get the needed ZZ critical value and then create a 99%99 \% Z confidence interval for the average final grade for the population by hand. Choose the correct statement below. a. The test is significant at the 1%1 \% significance level because 62.75 is inside the confidence interval you created. b. The test is not significant at the 1%1 \% significance level because 63.25 is inside the confidence interval you created. c. The test is not significant at the 1%1 \% significance level because 62.75 is inside the confidence interval you created. d. The test is significant at the 1%1 \% significance level because 62.75 is outside the confidence interval you created.

Studdy Solution

STEP 1

What is this asking? We're checking if the average grade of all students in this course is different from 62.75, using info from 100 students and knowing how spread out all the grades usually are. Watch out! Don't mix up the sample mean with the population mean!
Also, remember that being inside the confidence interval means things are *not* significantly different.

STEP 2

1. Find the Z critical value
2. Calculate the margin of error
3. Build the confidence interval
4. Check if 62.75 falls within the interval

STEP 3

We want a **99%** confidence interval, which means we have (100%99%)/2=0.5%(100\% - 99\%) / 2 = 0.5\% on each tail of our normal distribution.
So, we're looking for the Z-score that leaves **0.5%** in each tail, or **0.995** in the middle!

STEP 4

Using a Z-table or a calculator, we find that the critical Z-value for a 99% confidence interval is **2.576**.
This means that 99% of the data falls within 2.576 standard deviations of the mean.

STEP 5

The margin of error tells us how much "wiggle room" we have around our sample mean.
It's calculated by multiplying the **critical Z-value** by the **standard error**.

STEP 6

The **standard error** is the population standard deviation divided by the square root of the sample size.
In our case, the population standard deviation is **3.4**, and our sample size is **100**.
So, the standard error is 3.4/100=3.4/10=0.343.4 / \sqrt{100} = 3.4 / 10 = \textbf{0.34}.

STEP 7

Now, we multiply our **critical Z-value (2.576)** by our **standard error (0.34)** to get the margin of error: 2.5760.34=0.8762.576 \cdot 0.34 = \textbf{0.876}.

STEP 8

To build the confidence interval, we take our **sample mean (63.25)** and add/subtract the **margin of error (0.876)**.

STEP 9

The lower bound of our interval is 63.250.876=62.3763.25 - 0.876 = \textbf{62.37}.
The upper bound is 63.25+0.876=64.1363.25 + 0.876 = \textbf{64.13}.

STEP 10

So, our 99% confidence interval is [62.37,64.13][\textbf{62.37}, \textbf{64.13}].
This means we're 99% confident that the *true* average grade for *all* students falls somewhere between 62.37 and 64.13.

STEP 11

Our confidence interval is [62.37,64.13][\textbf{62.37}, \textbf{64.13}].
We want to see if **62.75** is inside this range.

STEP 12

Since 62.37<62.75<64.1362.37 < 62.75 < 64.13, the value **62.75** *is* inside our confidence interval!

STEP 13

Since 62.75 is *inside* our 99% confidence interval, we're not confident that the true average is different from 62.75.
The correct answer is (c).

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