If 2/3 of the 600 seeds sprouted, how many seeds did not sprout?

Mathematics

1 answer:

kupik [55]3 years ago

8 0

Answer:

200 seeds

Step-by-step explanation:

If 2/3 of the seeds sprouted, then (1-2/3) 1/3 of the seeds did not sprout.

1/3 * 600 = 600/3 = 200

200 seeds did not sprout

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National data indicates that 35% of households own a desktop computer. In a random sample of 570 households, 40% owned a deskt

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Answer:

Yes, this provide enough evidence to show a difference in the proportion of households that own a desktop.

Step-by-step explanation:

We are given that National data indicates that 35% of households own a desktop computer.

In a random sample of 570 households, 40% owned a desktop computer.

Let p = population proportion of households who own a desktop computer

SO, Null Hypothesis, $H_0$ : p = 25% {means that 35% of households own a desktop computer}

Alternate Hypothesis, $H_A$ : p $\neq$ 25% {means that % of households who own a desktop computer is different from 35%}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of 570 households who owned a desktop computer = 40%

n = sample of households = 570

So, test statistics = $\frac{0.40-0.35}{{\sqrt{\frac{0.40(1-0.40)}{570} } } } }$

= 2.437

Since, in the question we are not given with the level of significance at which to test out hypothesis so we assume it to be 5%. Now at 5% significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics doesn't lies within the range of critical values of z 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 % of households who own a desktop computer is different from 35%.

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