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3 years ago

Sarah used her calculator to work out the value of a number y .

Mathematics

1 answer:

mars1129 [50]3 years ago

5 0

Answer:

$7.75 \: \leqslant y \: < \: 7.85$

Step-by-step explanation:

The calculator approximated the value of y at 7.8.

We need to complete the error interval for y.

In other words, we calculate the upper and lower bound for y.

The level level of accuracy is to the nearest tenth.

To find the lower bound, we subtract half the degree of accuracy to get:

$y \geqslant 7.8 - 0.05 \\ y \geqslant 7.75$

To find the upper bound, add half the level of accuracy to get:

$y \: < \: 7.8 + 0.05 \\ y \: < \: 7.85$

Therefore the error interval is

$7.75 \: \leqslant y \: < \: 7.85$

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Answer:

t(h)= (h-200)/-4.9

The time to reach a height of 50 meters is 30.6 seconds

Step-by-step explanation:

To express t as a function of height (h) we must transform the equation below:

h=200-4.9t

h-200=-4.9t

(h-200)/-4.9=t

t(h)= (h-200)/-4.9

To find the time to reach a height of 50 meters, we must replace this value on the new equation:

t(h)=50-200/-4.9

t(h)= 30.612

The time to reach a height of 50 meters is 30.6 seconds

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Juli2301 [7.4K]

Answer:

$z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124$

Null hypothesis: $p\geq 0.8$

Alternative hypothesis: $p < 0.8$

Since is a left tailed test the p value would be:

$p_v =P(Z$

So the p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis: $p\leq 0.8$

Alternative hypothesis: $p > 0.8$

$p_v =P(Z>2.124)=0.013$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation

n=110 represent the random sample taken

X=97 represent the students who completed the modules before class

$\hat p=\frac{97}{110}=0.882$ estimated proportion of students who completed the modules before class

$p_o=0.8$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v{/tex} represent the p value (variable of interest) 2) Concepts and formulas to use We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.8 or 80%: Null hypothesis:[tex]p\geq 0.8$

Alternative hypothesis: $p < 0.8$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level assumed $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(Z$

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis: $p\leq 0.8$

Alternative hypothesis: $p > 0.8$

$p_v =P(Z>2.124)=0.013$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

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