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3 years ago

Please help me!

Mathematics

2 answers:

slava [35]3 years ago

8 0

1/6·30=5
so your answer is b)5

umka2103 [35]3 years ago

7 0

Answer:

5 times a 6 can be expected in 30 times.

Step-by-step explanation:

A die is rolled 30 times.

We have to find how many time a 6 can be expected in 30 times.

Probability of getting 6 in 1 time = $\frac{1}{6}$

⇒ Probability of getting 6 in 30 time = $\frac{1}{6}\times30$

= 5 times

Therefore, 5 times a 6 can be expected in 30 times.

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Question 5. We have 5-year statistics of average amount of wheat crop (tons) harvested from 1 km2 per year, the results are as f

tino4ka555 [31]

Answer:

a) $t=\frac{524.6-550}{\frac{27.682}{\sqrt{5}}}=-2.052$

$p_v =P(t_{(4)}$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

We can say that at 5% of significance the true mean is not significantly different from 550.

b) The 95% confidence interval would be given (490.184;559.016).

We are confident at 95% that the true mean is between (490.184;559.016).

Step-by-step explanation:

Part a

Data given and notation

560, 525, 496, 543, 499

The mean and sample deviation can be calculated from the following formulas:

$\bar X =\frac{\sum_{i=1}^n x_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)}{n-1}}$

$\bar X=524.6$ represent the sample mean

$s=27.682$ represent the sample standard deviation

$n=5$ sample size

$\mu_o =550$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the population mean is equal to 550, the system of hypothesis are:

Null hypothesis: $\mu = 550$

Alternative hypothesis: $\mu \neq 550$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{524.6-550}{\frac{27.682}{\sqrt{5}}}=-2.052$

P-value

We need to calculate the degrees of freedom first given by:

$df=n-1=5-1=4$

Since is a one-side left tailed test the p value would given by:

$p_v =P(t_{(4)}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

We can say that at 5% of significance the true mean is not significantly different from 550.

Part b

The confidence interval for a proportion is given by this formula

$\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}$

The degrees of freedom are 4 from part a

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.