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alekssr [168]
3 years ago
7

Understanding how a budget is created will help you develop good financial decision making skills. This is known as: Question 3

options: Financial expertise Financial literacy Accounting Math
Mathematics
1 answer:
Andreas93 [3]3 years ago
3 0

Answer:

True mate

Step-by-step explanation:

You might be interested in
A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. Assume that the distr
Marianna [84]

Answer:

(a) Decision rule for 0.01 significance level is that we will reject our null hypothesis if the test statistics does not lie between t = -2.651 and t = 2.651.

(b) The value of t test statistics is 1.890.

(c) We conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) P-value of the test statistics is 0.0662.

Step-by-step explanation:

We are given that a recent study focused on the number of times men and women who live alone buy take-out dinner in a month.

Also, following information is given below;

Statistic : Men      Women

The sample mean : 24.51      22.69

Sample standard deviation : 4.48    3.86

Sample size : 35    40

<em>Let </em>\mu_1<em> = mean number of times men order take-out dinners in a month.</em>

<em />\mu_2<em> = mean number of times women order take-out dinners in a month</em>

(a) So, Null Hypothesis, H_0 : \mu_1-\mu_2 = 0     {means that there is no difference in the mean number of times men and women order take-out dinners in a month}

Alternate Hypothesis, H_A : \mu_1-\mu_2\neq 0     {means that there is difference in the mean number of times men and women order take-out dinners in a month}

The test statistics that would be used here <u>Two-sample t test statistics</u> as we don't know about the population standard deviation;

                      T.S. =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } }  ~ t__n_1_-_n_2_-_2

where, \bar X_1 = sample mean for men = 24.51

\bar X_2 = sample mean for women = 22.69

s_1 = sample standard deviation for men = 4.48

s_2 = sample standard deviation for women = 3.86

n_1 = sample of men = 35

n_2 = sample of women = 40

Also,  s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2}  }{n_1+n_2-2} }  =  \sqrt{\frac{(35-1)\times 4.48^{2}+(40-1)\times 3.86^{2}  }{35+40-2} } = 4.16

So, <u>test statistics</u>  =  \frac{(24.51-22.69)-(0)}{4.16 \sqrt{\frac{1}{35}+\frac{1}{40}  } }  ~ t_7_3

                              =  1.890

(b) The value of t test statistics is 1.890.

(c) Now, at 0.01 significance level the t table gives critical values of -2.651 and 2.651 at 73 degree of freedom for two-tailed test.

Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) Now, the P-value of the test statistics is given by;

                     P-value = P( t_7_3 > 1.89) = 0.0331

So, P-value for two tailed test is = 2 \times 0.0331 = <u>0.0662</u>

4 0
3 years ago
What is the distance between -93 and 114 on a number line???
Alborosie

Answer:

207

Step-by-step explanation:

6 0
3 years ago
Relationship B has a lesser rate than Relationship A. The graph represents Relationship A.
Zinaida [17]
Given that Relationship B has a lesser rate than Relationship A and that the graph representing Relationship A is a f<span><span>irst-quadrant graph showing a ray from the origin through the points (2, 3) and (4, 6) where the horizontal axis label is Time in weeks and the vertical axis label is Plant growth in inches.</span>

The rate of relationship A is given by the slope of the graph as follows:

slope= \frac{6-3}{4-2} = \frac{3}{2} =1.5

To obtain which table could represent Relationship B, we check the slopes of the tables and see which has a lesser slope.

For table A.
Time (weeks) 3 6 8 10
Plant growth (in.) 2.25 4.5 6 7.5

slope= \frac{4.5-2.25}{6-3} = \frac{2.25}{3} =0.75

For table B.
Time (weeks) 3 6 8 10
Plant growth (in.) 4.8 9.6 12.8 16

</span><span><span>slope= \frac{9.6-4.8}{6-3} = \frac{4.8}{3} =1.6

</span> For tabe C.
Time (weeks) 3 4 6 9
Plant growth (in.) 5.4 7.2 10.8 16.2

</span><span>slope= \frac{7.2-5.4}{4-3} = \frac{1.8}{1} =1.8

For table D.
Time (weeks) 3 4 6 9
Plant growth (in.) 6.3 8.4 12.6 18.9</span>

<span>slope= \frac{8.4-6.3}{4-3} = \frac{2.1}{1} =2.1</span>

Therefore, the table that could represent Relationship B is table A.
7 0
3 years ago
Read 2 more answers
A boy jogs 250 meters in 110 seconds. What is his average speed in m/s?
babunello [35]
His average speed is 2.27 meters per second
8 0
3 years ago
5x70= 5 x_____tens <br> = ______tens =______? <br> Use place value to find the product.
zhuklara [117]
5 × 70 = 5 × 7 tens = 350

WHY?

1 ten = 10 ones

70 ÷ 10 = 7 

So 7 tens. 

Hope I helped ya! 

5 0
2 years ago
Read 2 more answers
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