Question

A security part that most 18% of homeowners have a home security and in words and in symbols. Then der whether the hypothes for this is whered wealed. Explain your reasoning Suase the hypothesis in words and in symbole Choose the correct answer OA. The null bypothesis expressed in words is, he proportion of a homeowners who own homestymo 10 The rulyphesis is expressed symbolcally, 'H D#0.18. OB. The hypothesis expressed in words is, the proportion of homeowners who can meet amis 0:18. The thesis expressed by 2018 OC. The nullypothesis essed in words in the proportion of those who come se om 16' The ruly is bly, 0:18 OD. The possessed in words, the proportion of the ones who owns a home security mom is most 0:18. The lysed bolcally, OA. The skemave hypothesis pressed in words in the portion of all homeowners who own home scurt am 0.18The strative hypothesis is pressed symbolically as O. The stative hypothesis ed in words is the proportion of al homeowners who own a home out miss than 0.18. The strive hypothesis embolically, 0.18 OC. The womalive hypothesis expressed in words, the proportion of all homeowners who own home culty is not guato. The missed caly, 0.18 OD. The shmative hypothesis expressed in words, the proportion of homewers who own a one scurt is more than 0.18. The motive hypothesis is expressed symbolcally as, 'H D018 The hypothesis tot contains

Answer 1

Null hypothesis

Ans D. H0<= 0.18

Alternative hypothesis

Ans D. H1> 0.18

The hypothesis test is left sided because alternative hypothesis contains <.

Frequency Distribution is Ans B.

The provided sample means are shown below: X = 2400 X2 = 2308 Also, the provided sample standard deviations are: $1 175 S2 = 53 14 and n2 = and the sample sizes are ni = 11. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: M1 = U2 Ha: U1 > M2 This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region Based on the information provided, the significance level is a = 0.1, and the degrees of freedom are df = 15.93. In fact, the degrees of freedom are computed as follows, assuming that the population variances are unequal: Hence, it is found that the critical value for this right-tailed test is tc 1.337, for a = 0.1 and df = 15.93. = The rejection region for this right-tailed test is R= {t: t > 1.337). (3) Test Statistics Since it is assumed that the population variances are unequal, the t-statistic is computed as follows: t = X1 - X2 s. + ni n2 2400 – 2308 = 1.861 1752 14 + 532 11

(4) Decision about the null hypothesis Since it is observed that t = 1.861 > tc = 1.337, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p= 0.0406, and since p 0.0406 < 0.1, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean Mi is greater than 42, at the 0.1 significance level. Confidence Interval The 90% confidence interval is 5.686 < M1 – M2 < 178.314.
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