Answer: No , at 0.05 level of significance , we have sufficient evidence to reject the claim that LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average.

Step-by-step explanation:

Let $\mu$ denotes the average hours of sleep per night.

As per given , we have

$H_0:\mu=7\\H_a:\mu$

, since $H_a$ is left-tailed and population standard deviation is unknown, so the test is a left-tailed t -test.

Also , it is given that ,

Sample size : n= 22

Sample mean : $\overline{x}=7.24$

Sample standard deviation : s= 1.93

Test statistic : $t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}$

i.e. $t=\dfrac{7.24-7}{\dfrac{1.93}{\sqrt{22}}}\approx0.58$

For significance level $\alpha=0.05$ and degree of freedom 21 (df=n-1),

Critical t-value for left-tailed test= $t_{\alpha, df}=t_{0.05,21}=- 1.7207$

Decision : Since the test statistic value (0.58) > critical value 1.7207, it means we are failed to reject the null hypothesis .

[Note : When $|t_{cal}|>|t_{cri}|$ , then we accept the null hypothesis.]

Conclusion: We have sufficient evidence to reject the claim that LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average.

6 0

2 years ago

Solve for y -2(x + 3 y) =18

LUCKY_DIMON [66]

Answer:

y= -3y - 1/3x

Step-by-step explanation:

-2(x + 3y) = 18 Distribute the -2 to the x + 3y

-2x - 6y = 18

+2x +2x Add the 2x to both sides

-6y = 18 + 2x Divide both sides by -6

y= -3y - 1/3x

3 0

3 years ago