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3 years ago

Convert 0.5 to a percent. A. 0.5% B. 50% C. 500% D. 5%

Mathematics

2 answers:

Sidana [21]3 years ago

5 0

The answer is 50% you move the decimal to the right twice.

jok3333 [9.3K]3 years ago

3 0

The answer is 50% because percent means per 100, so: Multiply the Decimal Number by 100.

Hoped this helped.

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3 years ago

A Statistics professor has observed that for several years students score an average of 114 points out of 150 on the semester ex

kirill [66]

Answer:

Reject $H_0$ . The change is statistically significant. The software does appear to improve exam scores.

Step-by-step explanation:

We are given that a Statistics professor has observed that for several years students score an average of 114 points out of 150 on the semester exam. The software is expensive, and the salesman offers to let the professor use it for a semester to see if the scores on the final exam increase significantly.

In the trial course that used this software, 218 students scored an average of 117 points on the final with a standard deviation of 8.7 points.

Let $\mu$ = mean scores on the final exam.

SO, Null Hypothesis, $H_0$ : $\mu \leq$ 114 points {means that the mean scores on the final exam does not increases after using software}

Alternate Hypothesis, $H_A$ : $\mu$ > 114 points {means that the mean scores on the final exam increase significantly after using software}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average points = 117 points

s = sample standard deviation = 8.7 points

n = sample of students = 218

So, test statistics = $\frac{117-114}{\frac{8.7}{\sqrt{218} } }$ ~ $t_2_1_7$

= 5.091

Now, P-value of the test statistics is given by the following formula;

P-value = P( $t_2_1_7$ > 5.091) = Less than 0.05%

Since, in the question we are not given with the level of significance at which hypothesis can be tested, so we assume it to be 5%. Now at 5% significance level, the t table gives critical value of 1.645 at 217 degree of freedom for right-tailed test. Since our test statistics is higher than the critical value of t as 5.091 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the change is statistically significant. The software does appear to improve exam scores.

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4 years ago