Is there a doctor in the house? A market research firm reported the mean annual earnings of all family practitioners in the United States was $178,258. A random sample of 52 family practitioners in Los Angeles had mean earnings of x$193,130 with a standard deviation of $42,047. Do the data provide sufficient evidence to conclude that the mean salary for family practitioners in Los Angeles is greater than the national average? Use the a 0.10 level of significance and the critical value method with the table. (a) State the appropriate null and alternate hypotheses. (b) Compute the value of the test statistic. ola (c) State a conclusion. Use the a = 0.10 level of significance.

Question
Is there a doctor in the house? A market research firm reported the mean annual earnings of all family practitioners in the
United States was $178,258. A random sample of 52 family practitioners in Los Angeles had mean earnings of x$193,130 with
a standard deviation of $42,047. Do the data provide sufficient evidence to conclude that the mean salary for family practitioners
in Los Angeles is greater than the national average? Use the a 0.10 level of significance and the critical value method with the
table.
(a) State the appropriate null and alternate hypotheses.
(b) Compute the value of the test statistic.
ola
(c) State a conclusion. Use the a
= 0.10 level of significance.

Image Transcription

Is there a doctor in the house? A market research firm reported the mean annual earnings of all family practitioners in the United States was $178,258. A random sample of 52 family practitioners in Los Angeles had mean earnings of x$193,130 with a standard deviation of $42,047. Do the data provide sufficient evidence to conclude that the mean salary for family practitioners in Los Angeles is greater than the national average? Use the a 0.10 level of significance and the critical value method with the table. (a) State the appropriate null and alternate hypotheses. (b) Compute the value of the test statistic. ola (c) State a conclusion. Use the a = 0.10 level of significance.

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