Try our AP Stats unit 9 test. This unit is all about linear regression, a method of finding a line of best fit through two-dimensional data. An important topic in this unit is testing the significance of a slope in a regression line. We adopt a null hypothesis that the slope is zero (the x-variable does not affect the y-variable at all) and then see how improbable the data would be.

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Question 1 |

### What should be used to compute a confidence interval for the slope of a regression model?

A t-interval | |

A z-score | |

A p-value | |

A Chi-squared test |

Question 1 Explanation:

The correct answer is (A). The appropriate confidence interval for the slope of a regression model is a t-interval for the slope.

Question 2 |

### If $n$ = 1002 and I am calculating the 95% confidence interval for the slope of the regression line using my point estimate of 3 for the slope and 2.37 for the standard error, which of the following is true?

The best critical value to use is 1.960. | |

Since $n$ ≥ 30 it would be fine if the relationship seemed nonlinear. | |

The standard error (2.37) is less than the slope (3), which is why we don’t have evidence that $y$ varies with $x$. | |

The confidence interval will include zero, which is why we don’t have evidence that $y$ varies with $x$. |

Question 2 Explanation:

The correct answer is (D). The best critical value you can get from the t-table provided on the AP Statistics exam is:

$1.962 \,$ for $\, df = n − 2 = 1002 - 2 = 1000$

1.960 is very close , but 1.962 is better in this case. The confidence interval will be:

$3 \pm 1.962*2.37 $ $ = (-1.650, 7.650)$

This includes zero. Which means that our 95% confidence interval contains zero, which means the $p$-value would not be below .05, so we wouldn’t reject the null hypothesis that the slope is zero.

Answer choice (B) is incorrect because we do need a linear relationship (or we at least need it not to look like a nonlinear relationship) between $x$ and $y$.

Answer choice (C) is incorrect — it is there to confuse you.

$1.962 \,$ for $\, df = n − 2 = 1002 - 2 = 1000$

1.960 is very close , but 1.962 is better in this case. The confidence interval will be:

$3 \pm 1.962*2.37 $ $ = (-1.650, 7.650)$

This includes zero. Which means that our 95% confidence interval contains zero, which means the $p$-value would not be below .05, so we wouldn’t reject the null hypothesis that the slope is zero.

Answer choice (B) is incorrect because we do need a linear relationship (or we at least need it not to look like a nonlinear relationship) between $x$ and $y$.

Answer choice (C) is incorrect — it is there to confuse you.

Question 3 |

### The slope of a given regression line of a sample size of 30 is 1.5, with a sample standard deviation s of 0.3 and a sample standard deviation of the $x$-values of 0.25. What is the margin of error for a critical value of 2.467?

0.382 | |

0.540 | |

0.550 | |

0.559 |

Question 3 Explanation:

The correct answer is (C). The margin of error for the slope is $t∗SE$, the standard error, which can be calculated by:

$ \dfrac{s}{s_x\sqrt{n-1}} = \dfrac{0.3}{0.25\sqrt{29}} ≈ 0.223$

$2.467*0.223≈0.550$

$ \dfrac{s}{s_x\sqrt{n-1}} = \dfrac{0.3}{0.25\sqrt{29}} ≈ 0.223$

$2.467*0.223≈0.550$

Question 4 |

### Which of these is the correct calculation of the confidence interval for the slope of the least squares regression line with a confidence interval of 95%, a sample size of 24, a slope of 0.386, and a standard error of sample slope of 0.082? Use the excerpt from the table below to find t*.

## df |
## Tail Probability p=0.05 |
## Tail Probability p=0.025 |

## 22 |
## 1.717 |
## 2.074 |

## 23 |
## 1.714 |
## 2.069 |

## 24 |
## 1.711 |
## 2.064 |

## 25 |
## 1.708 |
## 2.060 |

## 26 |
## 1.706 |
## 2.056 |

0.386 ± 2.074(0.082) | |

0.386 ± 2.064(0.082) | |

0.386 ± 2.056(0.082) | |

0.082 ± 2.074(0.386) |

Question 4 Explanation:

The correct answer is (A). The interval estimate is:

$b \pm t*(SE_b)$

In this case, $b=0.386, SE_b=0.082$, and $\, t* = 2.074 \,$ $ (p=0.025 \,$ with $\, df \; n − 2= 24 − 2 = 22)$.

The correct set-up for the calculation is:

$0.386 ± 2.074(0.082)$

$b \pm t*(SE_b)$

In this case, $b=0.386, SE_b=0.082$, and $\, t* = 2.074 \,$ $ (p=0.025 \,$ with $\, df \; n − 2= 24 − 2 = 22)$.

The correct set-up for the calculation is:

$0.386 ± 2.074(0.082)$

Question 5 |

### Which of the following describes the effects of sample size on the width of a confidence interval for the slope of a regression model?

As sample size increases, the width of the confidence interval increases. | |

As sample size increases, the width of the confidence interval decreases. | |

As sample size increases, the width of the confidence interval remains constant. | |

As sample size increases, the width of the confidence interval changes at random. |

Question 5 Explanation:

The correct answer is (B). All other factors remaining the same, the width of the confidence interval will decrease as the sample size increases. This is because the standard error has $n$ in the denominator, whether it’s a proportion, mean, difference in means, etc.

Question 6 |

### Which of the following would not be an appropriate null and alternative hypothesis for the slope of a regression model?

$H_0: β=0, H_a: β≠0$ | |

$H_0: β=0, H_a: β<0$ | |

$H_0: β=0, H_a: β>0$ | |

$H_0: β=0, H_a: β<0$ |

Question 6 Explanation:

The correct answer is (D). Appropriate alternative hypotheses can use <, >, or ≠ but cannot use ≤ or ≥ as this would overlap with the null hypothesis.

Question 7 |

### Which of the following is NOT a condition for the significance test for the slope of a regression model?

The true relationship between $x$ and $y$ is nonlinear. | |

The standard deviation for $y$ does not vary with $x$. | |

The samples are independent of each other. | |

The residuals seem normally distributed. |

Question 7 Explanation:

The correct answer is (A). The true relationship between $x$ and $y$ needs to be LINEAR (or at least look linear). All of the others are conditions for using linear regression to get the slope, and for performing the significance test of the slope.

Question 8 |

### What is the distribution of the slope of a regression model, assuming all conditions are satisfied and the null hypothesis is true?

Normal distribution | |

Z-distribution | |

T-distribution | |

Chi-squared distribution |

Question 8 Explanation:

The correct answer is (C). The distribution of the slope of a regression model, assuming all conditions are satisfied and the null hypothesis is true, is a t-distribution. That is why we use the t-table to get the confidence intervals for slopes.

Question 9 |

### How should one use the p-value to justify a claim based on the results of a significance test for the slope of a regression model if the significance level is $\alpha \,$?

If $p$-value = $\alpha$, accept the null hypothesis. | |

If $p$-value ≥ $\alpha$, reject the null hypothesis. | |

If $p$-value ≤ $\alpha$, reject the null hypothesis. | |

If $p$-value ≠ $\alpha$, accept the null hypothesis. |

Question 9 Explanation:

The correct answer is (C). If the $p$-value ≤ $\alpha$, then reject the null hypothesis; if the $p$-value > $\alpha$, accept the null hypothesis. $\alpha$ is the significance level, the threshold for rejecting the null hypothesis.

Question 10 |

### A student took a random sample of other students and found a linear relationship between the number of minutes spent in the library and ACT scores. A 95% confidence level for the slope of the regression line was (15, 55). The student wants to use this interval to test $H_0: β=0, H_a: β≠0$ at the α=0.05 level of significance. What is an appropriate conclusion?

Fail to reject $H_0$, and fail to conclude a linear relationship between library use and ACT scores. | |

Fail to reject $H_0$, suggesting a linear relationship between library use and ACT scores. | |

Reject $H_0$, and fail to conclude a linear relationship between library use and ACT scores. | |

Reject $H_0$, suggesting a linear relationship between library use and ACT scores. |

Question 10 Explanation:

The correct answer is (D). The null hypothesis says there is no relationship, that the slope of the regression line = 0. Because the confidence interval does not contain zero, the student can reject the null hypothesis and conclude that there is a linear relationship between library use and ACT scores.

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