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3 years ago

9.A bucket contains 25 kg of jelly beans.Cups containing 1.8 kg of jelly beans are measured out from the bucket.

Mathematics

1 answer:

lukranit [14]3 years ago

4 0

Answer:

a) 13 cups

b) 1 .6kg

Step-by-step explanation:

a) 25 / 1.8 = 13.89

13 cups

to check the answer:

13 cups x 1.8kg per cup = 23.4kg in total

we can use 23.4 to figure out b

b) 25 kg - 23.4kg = 1.6kg

1.6kg

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Answer:

We conclude that the true average weight loss for the vegan diet exceeds that for the control diet by more than 1 kg.

Step-by-step explanation:

We are given that an article reported on the results of an experiment in which half of the individuals in a group of 66 postmenopausal overweight women were randomly assigned to a particular vegan diet, and the other half received a diet based on National Cholesterol Education Program guidelines.

The sample mean decrease in body weight for those on the vegan diet was 6 kg, and the sample SD was 3.2, whereas, for those on the control diet, the sample mean weight loss and standard deviation were 3.8 and 2.4, respectively.

Let = true average weight loss for the vegan diet.

$\mu_2$ = true average weight loss for the control diet.

So, Null Hypothesis, : 1 kg {means that the true average weight loss for the vegan diet exceeds that for the control diet by less than or equal to 1 kg}

Alternate Hypothesis, : > 1 kg {means that the true average weight loss for the vegan diet exceeds that for the control diet by more than 1 kg}

The test statistics that will be used here is Two-sample t-test statistics because we don't know about population standard deviations;

T.S. = $\frac{(\bar X_1-\barX_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_+_n_2_-_2$

where, = sample mean weight loss for the vegan diet = 6 kg

= sample mean weight loss for the control diet = 3.8 kg

= sample standard deviation weight loss for the vegan diet = 3.2 kg

= sample standard deviation weight loss for the control diet = 2.4 kg

$n_1$ = sample of vegan diet women = 33

$n_2$ = sample of control diet women = 33

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(33-1)\times 3.2^{2}+(33-1)\times 2.4^{2} }{33+33-2} }$ = 2.83

So, the test statistics = $\frac{(6-3.8)-(1)}{2.83 \times \sqrt{\frac{1}{33}+\frac{1}{33} } }$ ~ $t_6_4$

= 1.722

The value of t-test statistics is 1.722.

Also, the P-value of the test statistics is given by;

P-value = P( $t_6_4$ > 1.722) = 0.0461 or 4.61%

Since the P-value of our test statistics is less than the level of significance as 0.0461 < 0.05, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average weight loss for the vegan diet exceeds that for the control diet by more than 1 kg.

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